מרכז תרגול

תרגול 1 היום: משפט לופיטל. • משפט דרבו. • תזכורת: משפט לופיטל: .(x0, b) פונקציות המוגדרות ב f, g ותהיינה x0 < b כך ש x0, b ∈ R תהיינה נניח כי . lim x→x+ 0 g (x) = ∞ או lim x→x+ 0 f (x) = lim x→x+ 0 g (x) = 0 .1 .(x0, b) גזירות ב f, g .2 .x ∈ (x0, b) לכל g′ (x) 6 = 0 .3 קיים במובן הרחב. lim x→x+ 0 f ′(x) g′(x) 4. הגבול קיים במובן הרחב, ובנוסף, lim x→x+ 0 f (x) g(x) אזי גם הגבול lim x→x+ 0 f (x) g (x) = lim x→x+ 0 f ′ (x) g′ (x) הערה: . ? ∞ , 0 0 כלל לופיטל תקף רק במקרים: 1

דוגמאות: חשבו כל אחד מהגבולות הבאים או הוכיחו שאינו קיים במובן הרחב: lim x→0 [ x · e 1 x ] .1 פתרון: נפריד לגבולות חד צדדיים. גבול מימין: lim x→0+ [ x · e 1 x ] = lim x→0+ e 1 x 1 x ? ∞ = L lim x→0+ e 1 x · (ZZ Z − 1 x2 ) ZZ Z − 1 x2 = ∞ כאשר lim x→0+ e 1 x = t= 1 x lim t→∞ et = ∞ גבול משמאל: lim x→0− x · e 1 x = AOL 0 · 0 = 0 כאשר lim x→0− e 1 x = t= 1 x lim t→−∞ et = 0 קיבלנו שהגבולות החד צדדיים שונים, לכן הגבול לא קיים במובן הרחב. 2

תזכורת: משפט דרבו: .[a, b] פונקציה המוגדרת ב f ותהי a < b כך ש a, b ∈ R תהיינה .f ′ (a) · f ′ (b) < 0 . נניח בנוסף כי[a, b] גזירה ב f נניח כי .f ′ (c) = 0 שעבורה a < c < b אזי קיימת משפט דרבו המורחב: .[a, b] פונקציה המוגדרת ב f ותהי a < b כך ש a, b ∈ R תהיינה .f ′ (b) ≤ r ≤ f ′ (a) או f ′ (a) ≤ r ≤ f ′ (b) וכי [a, b] גזירה ב f . נניח כיr ∈ R יהי .f ′ (c) = r שעבורה a ≤ c ≤ b אזי קיימת הוכחה: .(c = a) , אזי סיימנוf ′ (a) = r אם .(c = b) , אזי סיימנוf ′ (b) = r אם .f ′ (b) < r < f ′ (a) או f ′ (a) < r < f ′ (b) אחרת, .x ∈ [a, b] לכל g (x) = f (x) − rx נגדיר מתקיים x ∈ [a, b] מאשג"ז וכי לכל [a, b] גזירה ב g נשים לב כי g′ (x) = f ′ (x) − r , אזיf ′ (a) < r < f ′ (b) אם g′ (a) = f ′ (a) − r < 0 g′ (b) = f ′ (b) − r > 0 ,g′ (c) = 0 שעבורה a < c < b ולכן ממשפט דרבו קיימת g′ (a) · g′ (b) < 0 לכן כ

תרגיל: .{f ′ (x) : x ∈ R} ⊆ Q פונקציה גזירה. נניח כי f : R → R תהי .x ∈ R לכל f ′ (x) = q0 כך ש q0 ∈ Q היא קבועה, כלומר שקיים f ′ 1. הוכיחו כי פתרון: וכך ש x1 6 = x2 כך ש x1, x2 ∈ R לא קבועה, לכן קיימים f ′ נניח בשלילה ש .f ′ (x1) < f ′ (x2) כך ש r ∈ R \ Q קיים R \ Q מצפיפות f ′ (x1) < r < f ′ (x2) בסתירה f ′ (c) = r כך ש x2 ל x1 בין c לכן ממשפט דרבו המורחב קיימת נקודה .{f ′ (x) : x ∈ R} ⊆ Q לכך ש .x ∈ R לכל f ′ (x) = q0 כך ש q0 ∈ Q היא קבועה ולכן קיים f ′ לכן .x ∈ R לכל f (x) = q0 · x + r0 כך ש r0 ∈ R 2. הוכיחו כי קיים פתרון: .x ∈ R לכל g (x) = f (x) − q0x נגדיר את מתקיים x ∈ R גזירה מאשג"ז וכי לכל g נשים לב כי g′ (x) = f ′ (x) − q0 = q0 − q0 = 0 מתקיים x ∈ R כך שלכל r0 ∈ R , כלומר קייםR קבועה ב g ולכן g (x) = r0 מתקיים x ∈ R לכן לכל f (x) − q0x = r0 ⇒ f (x) = q0 · x + r0 5

הוכיחו או הפריכו: הוכיחו או הפריכו כל אחת מהטענות הבאות: , lim x→∞f ′ (x) = L פונקציה גזירה המקיימת f : R → R ותהי L > 0 1. יהי . lim x→∞f (x) = ∞ אזי פתרון: הטענה נכונה. נוכיח בעזרת לופיטל lim x→∞f (x) = lim x→∞x · f (x) x נחשב בנפרד את הגבול lim x→∞ f (x) x ? ∞ = L lim x→∞ f ′ (x) 1 = L נחזור לתרגיל ומאש"ג נקבל lim x→∞f (x) = lim x→∞x · f (x) x = AOL ∞ · L = ∞ קיים במובן הרחב. lim x→∞ f (x) x פונקציה גזירה. נניח כי f : R → R 2. תהי . lim x→∞ f (x) x = lim x→∞f ′ (x) קיים במובן הרחב ומתקיים lim x→∞f ′ (x) אזי פתרון: הטענה לא נכונה. .x ∈ R לכל f (x) = sin x דוגמא נגדית: נבחר נשים לב כי lim x→∞ f (x) x = lim x→∞ sin x x = lim x→∞ sin x · 1 x = B×V 0 ולכן x ∈ R לכל f ′ (x) = cos x אבל lim x→∞f ′ (x) = lim x→∞ cos x לא קיים במובן הרחב. 6

x ∈ R 3. הפונקציה המוגדרת לכל f (x) = { sin x x x 6 = 0 1 x = 0 גזירה. פתרון: הטענה נכונה. .x0 ∈ R יהי מאשג"ז. x0 6 = 0 גזירה לכל f נקבל x0 = 0 עבור f ′ (0) = lim x→0 f (x) − f (0) x − 0 = lim x→0 sin x x − 1 x = lim x→0 sin x − x x2 0 0 = L lim x→0 cos x − 1 2x 0 0 = L lim x→0 − sin x 2 = AOL 0 f ′ (x) = 0 כך ש [−1, 1] גזירה ולא קבועה בקטע f : [−1, 1] → R 4. קיימת פונקציה .x ∈ [−1, 0) ∪ (0, 1] לכל פתרון: הטענה לא נכונה. נניח בשלילה שקיימת פונקציה כזאת. ואז היא קבועה בקטע. x ∈ [−1, 1] לכל f ′ (x) = 0 , אחרתf ′ (0) 6 = 0 לכן , אזיf ′ (0) 6 = 0 ו f ′ (1) = 0 היות ש 0 = f ′ (1) < f ′ (0) 2 < f ′ (0) or f ′ (0) < f ′ (0) 2 < f ′ (1) = 0 כך ש 0 < c < 1 ולכן ממשפט דרבו המורחב קיימת נקודה f ′ (c) = f ′ (0) 2 6 = 0 סתירה. 7

תרגול 2 היום: סדרות. • תזכורת: .L ∈ R סדרה של מספרים ממשיים. יהי (an)∞ n=k ותהי 0 ≤ k ∈ Z יהי n ≥ N כך שלכל N ∈ N קיים > 0 אם לכל (an)∞ n=k הוא גבול של L נאמר כי מתקיים | an − L |< ונאמר שהסדרה מתכנסת. lim n→∞an = L במקרה זה נסמן משפט )הלמה של היינה(: .x0 פונקציה המוגדרת בסביבה מנוקבת של f . תהיL ∈ R ויהי x0 ∈ R תהי של מספרים ממשיים הנמצאת (xn)∞ n=1 , אזי לכל סדרה lim x→x0 f (x) = L אם . lim n→∞f (xn) = L מתקיים lim n→∞xn = x0 בסביבה המנוקבת, ושעבורה הערה: , אולם יש לעדכן את ההוכחהL = ±∞ ו x0 = ±∞ הלמה של היינה נכונה גם כאשר בהתאם. 1

דוגמאות: . חשבו את הגבול הבא, או הוכיחו שאינו קיים במובן הרחב0 < a ∈ R 1. יהי lim n→∞ n √a :פתרון .0 בסביבה המנוקבת של x לכל f (x) = ax נגדיר טבעי. n ≥ 1 לכל xn = 1 n נגדיר נשים לב ש lim n→∞xn = lim n→∞ 1 n = 0 ראינו באינפי 1 כי lim x→0f (x) = lim x→0ax = 1 ולכן מהלמה של היינה מתקיים 1 = lim n→∞f (xn) = lim n→∞f ( 1 n ) = lim n→∞a 1 n = lim n→∞ n √a 2

2. חשבו את הגבול הבא, או הוכיחו שאינו קיים במובן הרחב lim n→∞ n √n :פתרון .x ≥ 1 לכל f (x) = x 1 x נגדיר טבעי. n ≥ 1 לכל xn = n נגדיר נשים לב ש lim n→∞xn = lim n→∞n = ∞ נחשב את הגבול lim x→∞x 1 x = lim x→∞eln ( x 1 x ) = lim x→∞e ln x x = (∗) e0 = 1 נחשב את הגבול במעריך (∗) lim x→∞ ln x x ? ∞ = L lim x→∞ 1 x 1 = lim x→∞ 1 x = 0 ולכן מהלמה של היינה מתקיים 1 = lim n→∞f (xn) = lim n→∞f (n) = lim n→∞n 1 n 3

3. חשבו את הגבול הבא, או הוכיחו שאינו קיים במובן הרחב lim n→∞ [ n0.6 (n2 + 1)0.7 − n2] .x > 0 לכל f (x) = x0.7 רמז: היעזרו בפונקציה :פתרון .x > 0 לכל f (x) = x0.7 נגדיר .x > 0 לכל f ′ (x) = 0.7x−0.3 גזירה מאשג"ז ומתקיים f נשים לב ש כעת נחשב את הגבול lim n→∞ [ n0.6 (n2 + 1)0.7 − n2] = lim n→∞ n2 [ (n2 + 1)0.7 n1.4 − 1 ] = lim n→∞ n2 [( n2 + 1 n2 )0.7 − 1 ] = lim n→∞ n2 [( 1 + 1 n2 )0.7 − 1 ] = lim n→∞ (1 + 1 n2 )0.7 − 1 1 n2 = Heine xn= 1 n2 lim h→0 (1 + h)0.7 − 1 h = f ′ (1) = 0.7 הוכחה ללמה של היינה: סדרה של מספרים ממשיים הנמצאת בסביבה המנוקבת כך ש (xn)∞ n=1 תהי . > 0 ויהי lim n→∞xn = x0 0 <| x − x0 |< δ1 כך שלכל δ1 > 0 , קיים lim x→x0 f (x) = L מהיות הגבול מתקיים |f (x) − L| < (∗) מתקיים n ≥ N1 כך שלכל N1 ∈ N , קיים lim n→∞xn = x0 מהיות 0 < |xn − x0| < δ1 (∗∗) ,xn מתקיים עבור (∗) ולכן

תרגיל: המוגדרת באופן רקורסיבי על ידי (an)∞ n=1 נביט בסדרה { a1 = 1 4 an+1 = a2 n + 1 4 ∀n ∈ N ומצאו את ערכו. lim n→∞an הוכיחו כי קיים הגבול פתרון: .n ∈ N לכל 0 ≤ an < 1 2 ראשית נוכיח באינדוקציה ש .0 ≤ a1 = 1 4 < 1 2 בסיס האינדוקציה: .(0 ≤ an < 1 2 )כלומר n ונניח שהטענה מתקיימת עבור n ∈ N צעד האינדוקציה: יהי לכן an+1 = a2 n + 1 4 < 1 4 + 1 4 = 1 2 . 1 2 חסומה מלעיל על ידי (an)∞ n=1 ולכן הסדרה עולה. (an)∞ n=1 שנית נוכיח באינדוקציה ש .a2 = ( 1 4 )2 + 1 4 = 5 16 > 1 4 = a1 בסיס האינדוקציה: .(an+1 ≥ an )כלומר n ונניח שהטענה מתקיימת עבור n ∈ N צעד האינדוקציה: יהי ולכן n ∈ N לכל an ≥ 0 הוכחנו כי an+2 = a2 n+1 + 1 4 ≥ a2 n + 1 4 = an+1 עולה. (an)∞ n=1 הוכחנו כי הסדרה כך ש L ∈ R לכן ממשפט 1 מהרצאה 1 קיים L = lim n→∞an = sup {an : n ∈ N} . lim n→∞an+1 = L ולכן ממשפט 3 מהרצאה 1 מתקיים 5

תרגיל: המוגדרת באופן רקורסיבי על ידי (an)∞ n=1 נביט בסדרה c > 0 יהי { a1 = c an+1 = arctan (an) ∀n ∈ N ומצאו את ערכו. lim n→∞an הוכיחו כי קיים הגבול .x > 0 לכל arctan x < x רמז: פתרון: .n ∈ N לכל an > 0 ראשית נוכיח באינדוקציה ש .a1 = c > 0 בסיס האינדוקציה: .(an > 0 )כלומר n ונניח שהטענה מתקיימת עבור n ∈ N צעד האינדוקציה: יהי לכן an+1 = arctan (an) > 0 .(x > 0 לכל arctan x > 0 )נזכור ש .0 הוכחנו כי הסדרה חסומה מלרע על ידי .an+1 ≤ an מתקיים n ∈ N יורדת. נראה כי לכל (an)∞ n=1 שנית נוכיח ש .n ∈ N יהי an+1 = arctan (an) < an ולכן ניתן להשתמש ברמז. n ∈ N לכל an > 0 הוכחנו ש הוכחנו כי הסדרה יורדת. כך ש L ∈ R לכן ממשפט 2 מהרצאה 1 קיים L = lim n→∞an = inf {an : n ∈ N} . lim n→∞an+1 = L ולכן ממשפט 3 מהרצאה 1 מתקיים נשים לב כי lim n→∞an+1 = lim n→∞ arctan (an) כעת בעזרת הלמה של היינה ורציפות נקבל כי

ולכן נקבל כי L = arctan (L) .L = inf {an : n ∈ N} ≥ 0 , אזי0 היות שהסדרה חסומה מלרע על ידי שעבורו יש שוויון, L > 0 הוא פתרון של המשוואה ומהרמז לא יכול להיות L = 0 .L = 0 ולכן בהכרח L > arctan (L) כי בהכרח מתקיים . lim n→∞an = 0 כלומר קיבלנו כי 8

תרגול 3 היום: האינטגרל הלא מסוים. • תזכורת: f (x) ∫ f (x) dx xa,a 6 = −1 xa+1 a+1 + C 1 x ln | x | +C sin x − cos x + C cos x sin x + C ex ex + C ax, 1 6 = a > 0 ax ln a + C 1 cos2 x tan x + C 1 sin2 x − cot x + C 1 1+x2 arctan x + C 1 √1−x2 arcsin x + C − 1 √1−x2 arccos x + C לינאריות האינטגרל הלא מסוים: ∫ (af (x) + bg (x)) dx = ∫ af (x) dx + ∫ bg (x) dx = aF (x) + bG (x) + C 1

דוגמאות: .1 ∫ 1 x3 dx = ∫ x−3dx = x−2 −2 + C = − 1 2x2 + C .2 ∫ ( 2 x + ex ) dx = 2 ln | x | +ex + C תזכורת: הצבה:/החלפת משתנה ∫ g (F (x)) · f (x) dx = ∫ g (t) dt .dt = f (x) dx ו t = F (x) כאשר .3 ∫ 5 √x − 1dx = ∫ (x − 1)1 5 dx = t=x−1,dt=dx ∫ t1 5 dt = t6 5 6 5 + C = 5 6 (x − 1)6 5 + C הערה: בהצבה לינארית תמיד נקבל t = ax + b dt = adx dx = dt a 2

.4 ∫ ( 1 (1 + x)2 + 5 1 + x2 − 1 2x + 1 ) dx נפתור כל אינטגרל בנפרד ונשתמש בלינאריות האינטגרל ∫ 1 (1 + x)2 dx = t=1+x,dt=dx ∫ 1 t2 dt = ∫ t−2dt = t−1 −1 +C = −1 t +C = − 1 1 + x+C ∫ 5 1 + x2 dx = 5 arctan (x) + C ∫ 1 2x + 1dx = t=1+2x, dt 2 =dx ∫ 1 2tdt = 1 2 ln | t | +C = 1 2 ln | 2x + 1 | +C לכן סה"כ נקבל ∫ ( 1 (1 + x)2 + 5 1 + x2 − 1 2x + 1 ) dx = − 1 1 + x + 5 arctan(x) − 1 2 ln | 2x + 1 | +C .5 a, b > 0, ∫ 1 b + ax2 dx = 1 b ∫ 1 1 + a b x2 dx = 1 b ∫ 1 1 + (√a b x)2 dx = t=√a b x,dt=√a b dx 1 b ∫ 1 1 + t2 √ b adt = 1 b · √ b a·arctan (t)+C = 1 √ab · arctan (√a b x ) + C 3

הערה: ∫ f ′ (x) f (x) dx = ln | f (x) | +C ∫ f (x) f ′ (x) dx = f 2 (x) 2 + C .7 ∫ ( ex 1 + ex − sin x 3 + cos x ) dx נפתור כל אינטגרל בנפרד ונשתמש בלינאריות האינטגרל. נשים לב שהנגזרת של המכנה שווה למונה בשני המקרים. ∫ ex 1 + ex dx = ln | 1 + ex | +C ∫ − sin x 3 + cos xdx = ∫ − sin x 3 + cos xdx = ln | 3 + cos(x) | +C לכן סה"כ נקבל ∫ ( ex 1 + ex − sin x 3 + cos x ) dx = ln | 1 + ex | + ln | 3 + cos(x) | +C הערה: בשני האינטגרלים אפשר לוותר על הערך המוחלט, מכיוון שהביטויים בתוכם בהכרח חיוביים. .8 ∫ ex 1 + e2x dx = t=ex,dt=exdx ∫ 1 1 + t2 dt = arctan (t)+C = arctan (ex) + C 5

תזכורת: אינטגרציה בחלקים: ∫ u′ (x) v (x) dx = u (x) v (x) − ∫ u (x) v′ (x) dx .9 ∫ arcsin (x) dx נסמן u′ = 1, u = x, v = arcsin(x), v′ = 1 √1 − x2 נשתמש באינטגרציה בחלקים ונקבל ∫ arcsin (x) dx = x arcsin (x) − ∫ x √1 − x2 dx נפתור "בצד" את האינטגרל שקיבלנו ∫ x √1 − x2 dx = t=1−x2,dt=−2xdx ∫ 1 −2√tdt = ∫ t− 1 2 −2 dt = t1 2 1 2 · (−2) + C = − (1 − x2)1 2 + C לכן סה"כ נקבל ∫ arcsin(x)dx = x arcsin(x)− ( − (1 − x2)1 2 ) +C = x arcsin(x) + √1 − x2 + C 6

.10 ∫ ex cos (x) dx נסמן u′ = ex, u = ex, v = cos (x) , v′ = − sin (x) נשתמש באינטגרציה בחלקים ונקבל ∫ ex cos (x) dx = ex cos (x)− ∫ −ex sin (x) dx = ex cos (x)+ ∫ ex sin (x) dx נפתור "בצד" את האינטגרל שקיבלנו, שוב נסמן u′ = ex, u = ex, v = sin (x) , v′ = cos (x) ∫ ex sin (x) dx = ex sin (x) − ∫ ex cos (x) dx נשים לב שקיבלנו ∫ ex cos (x) dx = ex cos (x) + ex sin (x) − ∫ ex cos (x) dx נבודד את האינטגרל המבוקש ונקבל ∫ ex cos (x) dx = ex cos (x) + ex sin (x) 2 +C = ex(cos (x) + sin (x)) 2 + C 7

עוד דוגמאות: .11 ∫ sin3(x)dx = ∫ sin (x) · sin2(x)dx = ∫ sin (x) · (1 − cos 2 (x)) dx = ∫ sin (x) dx − ∫ sin (x) · cos2 (x) dx = נפתור כל אינטגרל בנפרד ונשתמש בלינאריות האינטגרל ∫ sin (x) dx = − cos (x) + C ∫ sin (x) · cos2(x)dx נשים לב שהנגזרת של הביטוי בחזקה שווה לביטוי השמאלי כפול מספר קבוע, נשתמש בהצבה t = cos (x) , dt = − sin (x) dx ⇒ sin (x) dx = −dt נציב לאינטגרל ונקבל ∫ −t2dt = −t3 3 + C = −cos 3 (x) 3 + C לכן סה"כ נקבל ∫ sin3 (x) dx = − cos (x)− ( −cos 3 (x) 3 ) +C = − cos (x) + cos 3 (x) 3 + C 8

Calculus II – Spring 2025-26 Homework 1 Solution 1. Compute each of the following limits or prove that it doesn’t exist in the extended sense (a) lim 𝑥→0 𝑥2 −𝑥 sin 𝑥 (𝑒𝑥 −1) ·ln(1+𝑥) . Solution: First note that lim 𝑥→0 𝑥 𝑒𝑥 − 1 0 0 = lim 𝑥→0 1 𝑒𝑥 AOL = 1 (★) and lim 𝑥→0 𝑥 − sin 𝑥 ln (1 + 𝑥) 0 0 = lim 𝑥→0 1 − cos 𝑥 1 1+𝑥 AOL = 0 (★★) It follows that lim 𝑥→0 𝑥2 − 𝑥 sin 𝑥 (𝑒𝑥 − 1) · ln (1 + 𝑥) = lim 𝑥→0 𝑥 (𝑥 − sin 𝑥) (𝑒𝑥 − 1) · ln (1 + 𝑥) = lim 𝑥→0 𝑥 𝑒𝑥 − 1 · 𝑥 − sin 𝑥 ln (1 + 𝑥) (★), (★★) = 0 (b) lim 𝑥→0+ 𝑥2 · ln(𝑥) . Solution: We have lim 𝑥→0+ 𝑥2 · ln(𝑥) = lim 𝑥→0+ ln(𝑥) 1 𝑥2 ? ∞ = lim 𝑥→0+ 1 𝑥 − 2 𝑥3 = lim 𝑥→0+ h − 𝑥 2 i AOL = 0 (c) lim 𝑥→0 tan( 𝑥) −𝑥 ln(1+𝑥) −𝑥 . Solution: we have lim 𝑥→0 tan(𝑥) − 𝑥 ln (

(d) lim 𝑥→∞ 𝑒2𝑥 − 𝑥 + 1 1 𝑥 . Solution: lim 𝑥→∞ 𝑒2𝑥 − 𝑥 + 1 1 𝑥 = lim 𝑥→∞𝑒ln (𝑒2𝑥 −𝑥+1) 1 𝑥 = lim 𝑥→∞𝑒 ln(𝑒2𝑥 −𝑥+1) 𝑥 = (★) Let us calculate the limit lim 𝑥→∞ ln 𝑒2𝑥 − 𝑥 + 1 𝑥 ? ∞ = lim 𝑥→∞ h 𝑒2𝑥 ·2−1 𝑒2𝑥 −𝑥+1 i 1 = lim 𝑥→∞ 𝑒2𝑥 · 2 − 1 𝑒2𝑥 − 𝑥 + 1 ? ∞ = lim 𝑥→∞ 4 · 𝑒2𝑥 2 · 𝑒2𝑥 − 1 ? ∞ = lim 𝑥→∞ 8 · 𝑒2𝑥 4 · 𝑒2𝑥 = lim 𝑥→∞2 = 2 We conclude that (★) = lim 𝑥→∞𝑒 ln [(𝑒2𝑥 −𝑥+1)] 𝑥 AOL = 𝑒2 (e) lim 𝑥→0 ( 𝑥+𝑒) 𝑥 −𝑒𝑥 𝑥2 . Solution: By AOD for every 𝑥 > −𝑒, (𝑥 + 𝑒)𝑥 is differentiable and ((𝑥 + 𝑒)𝑥 )′ = 𝑒𝑥 ln( 𝑥+𝑒) ′ = (𝑥 + 𝑒)𝑥 · ln (𝑥 + 𝑒) + 𝑥 𝑥 + 𝑒 Now lim 𝑥→0 (𝑥 + 𝑒)𝑥 − 𝑒𝑥 𝑥2 0 0 = lim 𝑥→0 (𝑥 + 𝑒)𝑥 · ln (𝑥 + 𝑒) + 𝑥 𝑥+𝑒 − 𝑒𝑥 2𝑥 0 0 = lim 𝑥→0 (𝑥 + 𝑒)𝑥 · ln (𝑥 + 𝑒) + 𝑥 𝑥+𝑒 2

2. Let 𝑥0 ∈ R and let 𝑓 be a function that is defined on a neighborhood of 𝑥0. Suppose that 𝑓 is differentiable at every 𝑥 in the neighborhood for which 𝑥 ≠ 𝑥0. Let 𝐿 ∈ R and suppose also that lim 𝑥→𝑥0 𝑓 ′ (𝑥) = 𝐿. (a) Give an example of a function satisfying the conditions above, and such that 𝑓 is not differentiable at 𝑥0. Solution: Consider the function defined by 𝑓 (𝑥) =    0 𝑥 ≠ 0 1 𝑥 = 0 for every 𝑥 ∈ R. We choose 𝑥0 = 0. Note that for every 𝑥 ≠ 0, 𝑓 is differentiable by AOD at 𝑥 and that for every 𝑥 ≠ 0 we have 𝑓 ′ (𝑥) = 0. It follows that lim 𝑥→0 𝑓 ′ (𝑥) = lim 𝑥→0 0 = 0 Note also that lim 𝑥→0 𝑓 (𝑥) = lim 𝑥→00 = 0 ≠ 1 = 𝑓 (0), which implies that 𝑓 is not continuous at 𝑥0 = 0 and therefore 𝑓 is not differentiable at 𝑥0 = 0. (b) Suppose

3. Let 𝑥0 ∈ R and let 𝑓 : R → R. (a) Suppose that 𝑓 is twice-differentiable at 𝑥0. Prove that lim ℎ→0 𝑓 (𝑥0 + ℎ) − 2 𝑓 (𝑥0) + 𝑓 (𝑥0 − ℎ) ℎ2 = 𝑓 ′′ (𝑥0) Solution: As 𝑓 is twice differentiable, then 𝑓 is continuous and therefore lim ℎ→0 𝑓 (𝑥0 + ℎ) = 𝑓 (𝑥0) and lim ℎ→0 𝑓 (𝑥0 − ℎ) = 𝑓 (𝑥0). It follows that lim ℎ→0 [ 𝑓 (𝑥0 + ℎ) − 2 𝑓 (𝑥0) + 𝑓 (𝑥0 − ℎ)] AOL = 0. Also, since 𝑓 is twice differentiable, 𝑓 ′ is differentiable in a neighborhood of 𝑥0. We now compute the limit starting from the left-hand side: lim ℎ→0 𝑓 (𝑥0 + ℎ) − 2 𝑓 (𝑥0) + 𝑓 (𝑥0 − ℎ) ℎ2 0 0 = lim ℎ→0 𝑓 ′ (𝑥0 + ℎ) + 𝑓 ′ (𝑥0 − ℎ) · (−1) 2ℎ = lim ℎ→0 𝑓 ′ (𝑥0 + ℎ) − 𝑓 ′ (𝑥0 − ℎ) 2ℎ = lim ℎ→0 𝑓 ′ (𝑥0 + ℎ) − 𝑓 ′ (𝑥0) + 𝑓 ′ (𝑥0) − 𝑓 ′ (𝑥0 − ℎ) 2ℎ = lim ℎ→0 𝑓 ′ (𝑥0 + ℎ) − 𝑓 ′ (𝑥0) 2

4. (a) Student A has computed lim 𝑥→1 𝑥3+3𝑥−4 𝑥2 −3𝑥+2 by using L’Hˆopital’s rule in the following way: lim 𝑥→1 𝑥3 + 3𝑥 − 4 𝑥2 − 3𝑥 + 2 = lim 𝑥→1 3𝑥2 + 3 2𝑥 − 3 = lim 𝑥→1 6𝑥 2 = 3 Explain student A’s mistake. Solution: Student A used L’Hˆopital’s rule but the conditions for L’Hˆopital’s rule don’t hold, as the limit lim 𝑥→1 3𝑥2+3 2𝑥−3 is not a limit of the form 0 0 nor of the form ? ∞ - it is a limit of the form 6 −1 . Student A could have used L’Hˆopital’s rule only on the first limit, but not on the second one. Indeed, the value of the limit is −6. (b) Student A has proved that every differentiable function 𝑓 : R → R must also be continuously differentiable in the following way: Let 𝑥0 ∈ R. Then 𝑓 ′ (𝑥0) def = lim 𝑥→𝑥0 𝑓 (𝑥) − 𝑓 (𝑥0) 𝑥 − 𝑥0 L’ˆopital 0 0 = l

5. Let 𝑓 be a function that is defined on (0, 2). Suppose that 𝑓 is differentiable. (a) Suppose that 𝑓 ′ (𝑥) · 𝑓 ′ 1 𝑥 < 0 for every 1 ≠ 𝑥 ∈ (0, 2). Prove that 𝑓 ′ (1) = 0. Solution: First note that as 𝑓 ′ (𝑥) · 𝑓 ′ 1 𝑥 < 0 for every 1 ≠ 𝑥 ∈ (0, 2), then 𝑓 ′ (𝑥) ≠ 0 for every 1 ≠ 𝑥 ∈ (0, 2). As 𝑓 ′ 2 3 · 𝑓 ′ 3 2 < 0, then by Darboux’s theorem there exists a 2 3 < 𝑐 < 3 2 for which 𝑓 ′ (𝑐) = 0. Thus, 𝑐 = 1. (b) Suppose that lim 𝑥→2− 𝑓 (𝑥) does not exist in the extended sense. Prove that there exists a 𝑐 ∈ (0, 2) for which 𝑓 ′ (𝑐) = 0. Solution: ATC that 𝑓 ′ (𝑥) ≠ 0 for every 0 < 𝑥 < 2. Suppose first that 𝑓 ′ (𝑥) > 0 for every 𝑥 ∈ (0, 2). Then 𝑓 is strictly increasing, and therefore lim 𝑥→2− 𝑓 (𝑥) exists in the extended sense and lim 𝑥→2− 𝑓 (𝑥

(c) Student X was kind of impressed when she saw the proof that sin(𝑥) < 𝑥 for every 0 < 𝑥 < 𝜋 2 in the first semester. Feeling powerful, she wanted to prove an inequality on her own: an inequality that would be both more impressive and more accurate. She asked her TA for help. Prove that there exists a 𝛿 > 0 such that for every 0 < 𝑥 < 𝛿 we have 𝑥 − 0.17𝑥3 < sin(𝑥) < 𝑥 − 0.16𝑥3 Solution: We compute lim 𝑥→0+ sin(𝑥) − 𝑥 𝑥3 0 0 = lim 𝑥→0+ cos(𝑥) − 1 3𝑥2 − 1 2 famous = − 1 6 As lim 𝑥→0+ sin( 𝑥) −𝑥 𝑥3 = − 1 6 , then there exists a 𝛿1 > 0 such that for every 0 < 𝑥 < 𝛿1 we have −0.17 < sin(𝑥) − 𝑥 𝑥3 < −0.16 (★) Choose 𝛿 = 𝛿1 and let 0 < 𝑥 < 𝛿. Then (★) hold for that 𝑥. Multiplying by 𝑥3 we obtain 𝑥 − 0.17𝑥3 < sin(𝑥) < 𝑥 − 0.16𝑥3. 7

6. Let 𝐿 ∈ R. (a) Let 𝑔1 : R → R. Suppose 𝑔1 is an odd function, i.e., 𝑔1 (−𝑥) = −𝑔1 (𝑥) for every 𝑥 ∈ R. Suppose in addition that lim 𝑥→∞𝑔1 (𝑥) = 𝐿. Prove that lim 𝑥→−∞𝑔1 (𝑥) = −𝐿. Solution: We compute lim 𝑥→−∞𝑔1 (𝑥) 𝑡=−𝑥 = lim 𝑡→∞𝑔1 (−𝑡) odd = lim 𝑡→∞ − 𝑔1 (𝑡) AOL = −𝐿 (b) Let 𝑔2 : R → R be a continuous function. Prove that if lim 𝑥→∞𝑔2 (𝑥) = lim 𝑥→−∞𝑔2 (𝑥) = 𝐿, then 𝑔2 has an extreme point in R. Solution: First, if 𝑔2 is constant then every point is extreme. We now assume 𝑔2 is not constant and consider the function defined by Ψ(𝑥) =    𝑔2 (tan(𝑥)) − 𝜋 2 < 𝑥 < 𝜋 2 𝐿 𝑥 = 𝜋 2 𝐿 𝑥 = − 𝜋 2 , ∀𝑥 ∈ h − 𝜋 2 , 𝜋 2 i Note that as 𝑔2 is continuous and as lim 𝑥→∞𝑔2 (𝑥) = lim 𝑥→−∞𝑔2 (𝑥) = 𝐿, then lim 𝑥→ 𝜋 2 − Ψ(𝑥) = lim

(c) Let 𝑓 : R → R be a differentiable function. Consider the function defined by 𝑔(𝑥) = 𝑓 (𝑥) − 𝑓 (−𝑥) for every 𝑥 ∈ R. Suppose that lim 𝑥→∞𝑔(𝑥) = 0. Conclude, using items (a) and (b), that there exists a 𝑐 ∈ R for which 𝑔′ (𝑐) = 0. Solution: We first show that 𝑔 is an odd function: let 𝑥 ∈ R. We note that 𝑔(−𝑥) = 𝑓 (−𝑥) − 𝑓 (−(−𝑥)) = 𝑓 (−𝑥) − 𝑓 (𝑥) = −( 𝑓 (𝑥) − 𝑓 (−𝑥)) = −𝑔(𝑥) Since we assumed lim 𝑥→∞ 𝑔(𝑥) = 0 then by item (a) we also obtain lim 𝑥→−∞𝑔(𝑥) = −0 = 0. As 𝑓 is differentiable, then by AOD 𝑔 is differentiable and therefore continuous and therefore by item (b) there exists an 𝑐 ∈ R such that 𝑐 is an extreme point. As 𝑔 is differentiable, then by Fermat’s theorem we have 𝑔′ (𝑐) = 0. (d) Prove that there exists a 𝑑 ∈ R for which 𝑓 ′ (�

Calculus II – Spring 2025-26 Homework 2 Solution 1. Define each of the following terms. Give a full and complete mathematical definition. If your definition includes a secondary term that was defined in the course, you must define it as well! You do not have to define any term that appears prior to the words ’define the term’. You are not required to re-define terms that were already defined. Let (𝑎𝑛)∞ 𝑛=1 be a sequence of real numbers. (a) Define the term: the sequence (𝑎𝑛)∞ 𝑛=1 is not bounded from below and not bounded from above, without using any negation word. Solution: We say that the sequence (𝑎𝑛)∞ 𝑛=1 is not bounded from below and not bounded from above if for every 𝐾1 ∈ R there exists an 𝑛1 ∈ N for which 𝑎𝑛1 < 𝐾1 and for every 𝐾2 ∈ R there exists an 𝑛2 ∈ N for whic

2. Let (𝑎𝑛)∞ 𝑛=1 be a sequence of real numbers. Prove each of the following statements. In items (a) and (b) prove using only the definitions. (a) For every 𝑘 ∈ N, the sequence (𝑎𝑛+𝑘 )∞ 𝑛=1 is convergent if and only if (𝑎𝑛)∞ 𝑛=1 is convergent, and in that case, lim 𝑛→∞𝑎𝑛+𝑘 = lim 𝑛→∞𝑎𝑛. Solution: Let 𝑘 ∈ N. We define (𝑏𝑛)∞ 𝑛=1 = (𝑎𝑛+𝑘 )∞ 𝑛=1. ⇐: As (𝑎𝑛)∞ 𝑛=1 is convergent, then there exists an 𝐿 ∈ R such that lim 𝑛→∞𝑎𝑛 = 𝐿. Let 𝜖 > 0. As lim 𝑛→∞𝑎𝑛 = 𝐿, then there exists an 𝑁1 ∈ N such that for every 𝑛 ≥ 𝑁1 we have |𝑎𝑛 − 𝐿| < 𝜖 (★) We choose 𝑁 = 𝑁1 . Let 𝑛 ≥ 𝑁, therefore 𝑛 + 𝑘 ≥ 𝑁1 and therefore (★) holds for 𝑛 + 𝑘. Then |𝑏𝑛 − 𝐿| = |𝑎𝑛+𝑘 − 𝐿| < 𝜖 Thus (𝑏𝑛)∞ 𝑛=𝑘 is convergent and lim 𝑛→∞𝑎𝑛 = lim 𝑛→∞𝑏𝑛 = 𝐿 . ⇒: As (𝑏𝑛)

(b) If (𝑎𝑛)∞ 𝑛=1 is increasing, then (𝑎𝑛)∞ 𝑛=1 converges in the extended sense and lim 𝑛→∞ 𝑎𝑛 = sup {𝑎𝑛 : 𝑛 ∈ N} Solution: First we assume that (𝑎𝑛)∞ 𝑛=1 is bounded from above. As (𝑎𝑛)∞ 𝑛=1 is bounded from above (and clearly not empty), then by the least upper bound theorem, there exists an ¯𝑠 ∈ R such that ¯𝑠 = sup {𝑎𝑛 | 𝑛 ∈ N}. Let 𝜖 > 0. By the 𝜖 property of ¯𝑠, there exists an 𝑛1 ≥ 1 for which ¯𝑠 − 𝜖 < 𝑎𝑛1 ≤ ¯𝑠. We choose 𝑁 ∈ N such that 𝑁 > 𝑛1 (by the Archimedean property there exists such N). let 𝑛 ≥ 𝑁. As (𝑎𝑛)∞ 𝑛=1 is increasing and 𝑛 > 𝑛1, we have ¯𝑠 − 𝜖 < 𝑎𝑛1 ≤ 𝑎𝑛 note that 𝑎𝑛 ≤ ¯𝑠 < ¯𝑠 + 𝜖, so ¯𝑠 − 𝜖 < 𝑎𝑛1 ≤ 𝑎𝑛 < ¯𝑠 + 𝜖 therefore |𝑎𝑛 − ¯𝑠| < 𝜖. We proved that (𝑎𝑛)∞ 𝑛=1 converges, and that lim 𝑛→∞𝑎𝑛 = ¯𝑠. Now

3. (a) Let 𝑎 ∈ R and let 𝑓 be a function that is defined on [𝑎, ∞). Assume that lim 𝑥→∞ 𝑓 (𝑥) = ∞. Prove that for every sequence (𝑥𝑛)∞ 𝑛=1 such that lim 𝑛→∞𝑥𝑛 = ∞, we have lim 𝑛→∞ 𝑓 (𝑥𝑛) = ∞. Solution: Let (𝑥𝑛)∞ 𝑛=1 be a sequence such that lim 𝑛→∞𝑥𝑛 = ∞. Let 𝐾 > 0. As lim 𝑥→∞ 𝑓 (𝑥) = ∞ then there exists an 𝑀1 > 0 such that for every 𝑥 > 𝑀1 we have 𝑓 (𝑥) > 𝐾 (★) As lim 𝑛→∞𝑥𝑛 = ∞ then there exists an 𝑁1 ∈ N such that for every 𝑛 ≥ 𝑁1 we have 𝑥𝑛 > 𝑀1 (★★) Choose 𝑁 = 𝑁1 and let 𝑛 ≥ 𝑁. Then (★★) holds for 𝑛 and therefore (★) holds for 𝑥𝑛. It follows that 𝑓 (𝑥𝑛) (★) > 𝐾 therefore lim 𝑛→∞ 𝑓 (𝑥𝑛) = ∞. (b) Compute the following limit: lim 𝑛→∞ 𝑛 + 𝑛2 ln 𝑛 − 𝑛2 ln(𝑛 + 1) Solution: We have lim 𝑛→∞ 𝑛 + 𝑛2 ln 𝑛 − 𝑛2 ln(𝑛 + 1) = lim

4. Compute the following limit as a function of 0 < 𝛼 < 1: lim 𝑛→∞ (𝑛 + 1)2𝛼 − 𝑛2 + 1 𝛼 Solution: lim 𝑛→∞ (𝑛 + 1)2𝛼 − (𝑛2 + 1) 𝛼 𝑥𝑛=𝑛 = lim 𝑥→∞ (𝑥 + 1)2𝛼 − (𝑥2 + 1) 𝛼 = lim 𝑥→∞ 𝑥2𝛼 " 1 + 1 𝑥 2𝛼 − 1 + 1 𝑥2 𝛼 # = lim 𝑥→∞ 1 + 1 𝑥 2𝛼 − 1 + 1 𝑥2 𝛼 𝑥−2𝛼 𝐿 = lim 𝑥→∞ − 2𝛼 𝑥2 1 + 1 𝑥 2𝛼−1 + 2𝛼 𝑥3 1 + 1 𝑥2 𝛼−1 −2𝛼𝑥−2𝛼−1 = lim 𝑥→∞ 𝑥−2 1 + 1 𝑥 2𝛼−1 − 𝑥−3 1 + 1 𝑥2 𝛼−1 𝑥−2𝛼−1 = lim 𝑥→∞ " 𝑥2𝛼−1 1 + 1 𝑥 2𝛼−1 − 𝑥2𝛼−2 1 + 1 𝑥2 𝛼−1# Now we consider three cases. Case 1: 0 < 𝛼 < 1 2 lim 𝑥→∞ " 𝑥2𝛼−1 1 + 1 𝑥 2𝛼−1 − 𝑥2𝛼−2 1 + 1 𝑥2 𝛼−1# AOL = 0 − 0 = 0 Case 2: 𝛼 = 1 2 lim 𝑥→∞ " 𝑥2𝛼−1 1 + 1 𝑥 2𝛼−1 − 𝑥2𝛼−2 1 + 1 𝑥2 𝛼−1# AOL = 1 − 0 = 0 Case 3: 𝛼 > 1 2 lim 𝑥→∞ " 𝑥2𝛼−1 1 + 1 𝑥 2𝛼−1 −

5. (a) Consider the sequence (𝑎𝑛)∞ 𝑛=1 defined by the recursive formula    𝑎𝑛+1 = 𝑎𝑛 + 1 𝑎𝑛 ∀𝑛 ∈ N 𝑎1 = 1 Prove that lim 𝑛→∞𝑎𝑛 = ∞. Solution: We first prove that ∀𝑛 ∈ N we have 𝑎𝑛 > 0, and therefore the sequence (𝑎𝑛)∞ 𝑛=1 is well-defined. We prove this claim by induction: Base case: 𝑎1 > 0. Induction step: Let 𝑛 ∈ N such that 𝑎𝑛 > 0. Then 𝑎𝑛+1 = 𝑎𝑛 + 1 𝑎𝑛 > 0. We now prove that the sequence (𝑎𝑛)∞ 𝑛=1 is increasing. Let 𝑛 ∈ N. Then 𝑎𝑛+1 = 𝑎𝑛 + 1 𝑎𝑛 > 𝑎𝑛 ATC that lim 𝑛→∞𝑎𝑛 ≠ ∞. As (𝑎𝑛)∞ 𝑛=1 is increasing, then there exists an 𝐿 ∈ R such that 𝐿 = lim 𝑛→∞𝑎𝑛 increasing = sup {𝑎𝑛 : 𝑛 ∈ N} . Note that 𝐿 upper bound ≥ 𝑎1 = 1 and therefore 𝐿 ≠ 0. It follows that 𝐿 = lim 𝑛→∞ 𝑎𝑛+1 = lim 𝑛→∞ 𝑎𝑛 + 1 𝑎𝑛 AOL = 𝐿 + 1 𝐿 Thus

(b) i. Let 0 ≤ 𝛼 ≤ 2. Prove that 𝛼 ≤ √2𝛼 ≤ 2. Solution: We first prove the right inequality. As 0 ≤ 𝛼 ≤ 2, then we have √𝛼 ≤ √2. Multiplying both sides of the inequality by √2 yields √2𝛼 ≤ √22 = 2. We now prove the left inequality. As 0 ≤ 𝛼 ≤ 2, then 𝛼2 ≤ 2𝛼. Taking the square root to both sides of the inequality yields 𝛼 ≤ √2𝛼. ii. Let (𝑎𝑛)∞ 𝑛=1 be the sequence of real numbers defined by the recursive formula    𝑎𝑛+1 = √2𝑎𝑛 ∀𝑛 ∈ N 𝑎1 = √2 Prove that the sequence (𝑎𝑛)∞ 𝑛=1 is convergent. Solution: We first prove that for every 𝑛 ∈ N we have 𝑎𝑛 ≥ 0, from which follows that the sequence (𝑎𝑛)∞ 𝑛=1 is well-defined. We prove that by using induction: Base case: we have 𝑎1 = √2 ≥ 0. Induction step: Let 𝑛 ∈ N and suppose that 𝑎𝑛 ≥ 0. Then 𝑎𝑛+1 =√︁ 2𝑎𝑛

2𝑎𝑛 AOL = √2𝐿 Thus 𝐿 = √2𝐿 and therefore 𝐿2 = 2𝐿 and therefore 𝐿 = 0 or 𝐿 = 2. As 𝐿 ≠ 0 then 𝐿 = 2. 7

6. Let 0 < 𝛽 < 𝛼 and let (𝑎𝑛)∞ 𝑛=1 and (𝑏𝑛)∞ 𝑛=1 be two sequences that are defined by the recursive formulas    𝑎𝑛+1 = 𝑎𝑛+𝑏𝑛 2 , 𝑏𝑛+1 = 2𝑎𝑛 𝑏𝑛 𝑎𝑛+𝑏𝑛 , ∀𝑛 ∈ N 𝑎1 = 𝛼, 𝑏1 = 𝛽 (a) Prove that (𝑎𝑛)∞ 𝑛=1 and (𝑏𝑛)∞ 𝑛=1 are well-defined, and prove that 𝑎𝑛+1 ≥ 𝑏𝑛+1 for every 𝑛 ∈ N. Solution: We first prove by induction that 𝑎𝑛, 𝑏𝑛 > 0 for every 𝑛 ∈ N. Base case: (𝑛 = 1). It is already given that 0 < 𝑏1 < 𝑎1. Induction step: (𝑛 ⇒ 𝑛 + 1.) Let 𝑛 ∈ N and suppose that 𝑎𝑛, 𝑏𝑛 > 0. Then 𝑎𝑛+1 = 𝑎𝑛+𝑏𝑛 2 > 0, 𝑏𝑛+1 = 2𝑎𝑛 𝑏𝑛 𝑎𝑛+𝑏𝑛 > 0. Thus (𝑎𝑛)∞ 𝑛=1 and (𝑏𝑛)∞ 𝑛=1 are well-defined. We prove now that 𝑎𝑛+1 ≥ 𝑏𝑛+1 for every 𝑛 ∈ N. Let 𝑛 ∈ N. As (𝑎𝑛 − 𝑏𝑛)2 ≥ 0 for every 𝑛 ∈ N, then 𝑎2 𝑛 − 2𝑎𝑛 𝑏𝑛 + 𝑏2 𝑛 ≥ 0 ⇒ 𝑎

We now show 𝐿1 = 𝐿2 and complete the proof: 𝐿1 = lim 𝑛→∞𝑎𝑛+1 = lim 𝑛→∞ 𝑎𝑛 + 𝑏𝑛 2 = 𝐿1 + 𝐿2 2 ⇒ 𝐿1 2 = 𝐿2 2 ⇒ 𝐿1 = 𝐿2. (c) Prove that 𝑎𝑛 𝑏𝑛 = 𝑎1 𝑏1 for every 𝑛 ∈ N and find the value of 𝐿. Solution: We prove by induction that 𝑎𝑛 𝑏𝑛 = 𝑎1 𝑏1 for every 𝑛 ∈ N. Base case: Clearly 𝑎1 𝑏1 = 𝑎1 𝑏1. Induction step: Let 𝑛 ∈ N. We assume that 𝑎𝑛 𝑏𝑛 = 𝑎1 𝑏1. Then 𝑎𝑛+1 𝑏𝑛+1 = 𝑎𝑛+𝑏𝑛 2 · 2𝑎𝑛 𝑏𝑛 𝑎𝑛+𝑏𝑛 = 𝑎𝑛 𝑏𝑛 = 𝑎1 𝑏1. Now we find the value of 𝐿. We note that lim 𝑛→∞ [𝑎1 𝑏1] = 𝑎1 𝑏1. By item (b) we obtain that lim 𝑛→∞ [𝑎𝑛 𝑏𝑛] = 𝐿2 Thus 𝐿2 = 𝑎1 𝑏1 ⇒ 𝐿 = ±√𝑎1 𝑏1 As 𝑎𝑛, 𝑏𝑛 > 0 for every 𝑛 ∈ N, then by the monotonicity of limits we have 𝐿 ≥ 0, thus 𝐿 = √𝑎1 𝑏1. 9

Calculus II – Spring 2025-26 Homework 3 Solution 1. Compute each of the following indefinite integrals: (a) ∫ 𝑒√𝑥 𝑑𝑥. Solution: We write ∫ 𝑒√𝑥 𝑑𝑥 𝑡=√𝑥→2√𝑥𝑑𝑡=𝑑 𝑥 = ∫ 𝑒𝑡 · 2𝑡 𝑑𝑡 = 𝑒𝑡 · 2𝑡 − ∫ 𝑒𝑡 · 2 𝑑𝑡 = 𝑒𝑡 · 2𝑡 − 𝑒𝑡 · 2 + 𝐶 = 2√𝑥𝑒√𝑥 − 2𝑒√𝑥 + 𝐶 (b) ∫ 𝑑 𝑥 sin2 ( 𝑥) cos2 ( 𝑥) . Hint. use the formula sin2 (𝑥) + cos2 (𝑥) = 1 for every 𝑥 ∈ R. Solution: We write ∫ 𝑑𝑥 sin2 (𝑥) cos2 (𝑥) formula = ∫ sin2 (𝑥) + cos2 (𝑥) sin2 (𝑥) cos2 (𝑥) 𝑑𝑥 = ∫ 𝑑𝑥 cos2 (𝑥) + ∫ 𝑑𝑥 sin2 (𝑥) = tan(𝑥) − cot(𝑥) + 𝐶 (c) ∫ 𝑥2 · 𝜆𝑒−𝜆𝑥 𝑑𝑥, where 𝜆 ≠ 0. Solution: We develop ∫ 𝑥2 · 𝜆𝑒−𝜆𝑥 𝑑𝑥 IBP = 𝜆 − 𝑥2𝑒−𝜆𝑥 𝜆 − ∫ − 2𝑥𝑒−𝜆𝑥 𝜆 𝑑𝑥 = 𝜆 − 𝑥2𝑒−𝜆𝑥 𝜆 + 2 𝜆 ∫ 𝑥𝑒−𝜆𝑥 𝑑𝑥 IBP = 𝜆 − 𝑥2𝑒−𝜆𝑥 𝜆 + 2 𝜆 − 𝑥𝑒−𝜆𝑥 𝜆 − ∫ 𝑒−�

The following questions are not in the exam material 2022 term A: 3a, 4b. 2022 term B: 5b, 5c. 2023 term A: 5. 2023 term B: 1b, 5a, 5b. 2024 term A: 3.1, 5.1. 2024 term B: 1.1, 2.1.3, 5.2.

2 Integration and differentiation R xα dx = xα+1 α+1 + C , (α , −1) (xα)′ = αxα−1 R 1 x dx = ln |x| + C (ln x)′ = 1 x R cos x dx = sin x + C (sin x)′ = cos x R sin x dx = − cos x + C (cos x)′ = − sin x R 1 cos2 x dx = tan x + C (tan x)′ = 1 cos2 x R ax dx = ax ln a + C (ax)′ = ln a · ax R ex dx = ex + C (ex)′ = ex R 1 1+x2 dx = arctan x + C (arctan x)′ = 1 1+x2 R 1 √1−x2 dx = arcsin x + C (arcsin x)′ = 1 √1−x2 R 1 − √1−x2 dx = arccos x + C (arccos x)′ = − 1 √1−x2 R 1 a2+x2 dx = 1 a arctan x a + C loga x ′ = 1 ln(a)·x R 1 √a2−x2 dx = arcsin x a + C R α · f (x) dx = α · R f (x) dx α · f (x) ′ = α · f ′(x) R f (x) ± g(x) dx = R f (x) dx ± R g(x) dx f (x) ± g(x) ′ = f ′(x) ± g′(x) R u′(x)v(x) dx = u(x)v(x) − R u(x)v′(x) dx [u(x) · v(x)]′ = u′(x)v(x) + u(x)v′(x) R g( f (x)) f ′(x) dx = R g(t

Infinitesimal Calculus 2, list of theorem for Quiz 2 1. Prove that the Dirichlet function is not integrable on any closed interval [𝑎, 𝑏] ⊆ R. 2. Let 𝑎, 𝑏 ∈ R such that 𝑎 < 𝑏 and let 𝑓 be a function that is bounded on [𝑎, 𝑏]. Suppose 𝑓 is integrable on [𝑎, 𝑏] Prove ∫ 𝑏 𝑎 𝑓 (𝑥) 𝑑𝑥 Δ ≤ ∫ 𝑏 𝑎 | 𝑓 (𝑥)| 𝑑𝑥. 3. Let 𝑎, 𝑏 ∈ R such that 𝑎 < 𝑏 and let 𝑓 be a function that is continuous on [𝑎, 𝑏]. Prove ∃𝑐 ∈ [𝑎, 𝑏] such that 𝑓 (𝑐) = 1 𝑏 − 𝑎 ∫ 𝑏 𝑎 𝑓 (𝑥) 𝑑𝑥. 4. Let 𝑎, 𝑏 ∈ R such that 𝑎 < 𝑏 and let 𝑓 be a function that is bounded on [𝑎, 𝑏]. Suppose 𝑓 is integrable on [𝑎, 𝑏] and consider the function defined by 𝐹 (𝑥) = ∫ 𝑥 𝑎 𝑓 (𝑡) 𝑑𝑡, ∀𝑥 ∈ [𝑎, 𝑏]. Prove that 𝐹 is continuous. 5. Let 𝑎, 𝑏 ∈ R such that 𝑎 < 𝑏 and let 𝑓 be a function that

Infinitesimal Calculus 2, list of theorem for Quiz 1 1. Let (𝑎𝑛)∞ 𝑛=1 be a sequence of real numbers. Let (𝑛𝑘 )∞ 𝑘=1 be a strictly increasing sequence of natural numbers. Suppose that the sequence (𝑎𝑛)∞ 𝑛=1 converges in the extended sense. Prove that the sequence (𝑎𝑛𝑘 )∞ 𝑘=1 converges in the exended sense and lim 𝑘→∞ 𝑎𝑛𝑘 = lim 𝑛→∞ 𝑎𝑛 2. Let (𝑎𝑛)∞ 𝑛=1 be a sequence of real numbers. Suppose (𝑎2𝑘 )∞ 𝑘=1 and (𝑎2𝑘−1)∞ 𝑘=1 both converge in the extended sense, and lim 𝑘→∞ 𝑎2𝑘 = lim 𝑘→∞ 𝑎2𝑘−1. Prove that the sequence (𝑎𝑛)∞ 𝑛=1 converges in the extended sense, and lim 𝑛→∞ 𝑎𝑛 = lim 𝑘→∞ 𝑎2𝑘 = lim 𝑘→∞ 𝑎2𝑘−1. 3. Let 𝑓 , 𝐹, 𝐺 be three functions defined on an interval 𝐼. Suppose that 𝐹, 𝐺 are anti-derivatives of 𝑓 . Prove ∃𝐶 ∈ R such that 𝐹 (𝑥) = 𝐺

Infinitesimal Calculus 2, Moed A 2022, Reichman university Solutions Lecturer: Dr. Yossi Shamai Instructions: 1. Duration: 3 hours. 2. Only a basic calculator and the attached formula sheets are permitted in the exam. Instructions: The following exam has 5 questions. You must answer exactly 4 questions. Each question is worth 25 credits. You must explain in full details all of your answers in all parts of the exam, and state all of the conditions for every theorem and/or claim. Unexplained answers will not receive any credit. It is forbidden to rely on any material that was not learned in the course. You must write your solutions only in the designated areas. Do not write at the back of the papers!

2 1. (25 points) (a) (13 points) Let (an)∞ n=1 be a sequence of real numbers. Suppose that the series ∑∞ n=1 an is convergent. i. (7 points) Prove that limn→∞ an = 0. ii. (6 points) Prove that the series ∑∞ n=1 an n2 converges absolutely. Solution: We have limn→∞ [ |an| n2 ] [ 1 n2 ] = limn→∞ |an| = 0, and as ∑∞ n=1 1 n2 is convergent then ∑∞ n=1 |an| n2 is convergent, that is, the series ∑∞ n=1 an n2 converges absolutely. (b) (12 points) Determine whether the following improper integral is convergent or divergent: ∫ 1 0 ex +x √ex −1 dx. Solution: Note that limx→0 ex −1 x = 1, which implies that there exists a 0 < δ1 < 1 for which for every 0 < x < δ1 we have ex −1 x > 1 2 . Thus, ∫ 1 0 ex + x √ex − 1 dx = ∫ δ1 0 ex + x √ex − 1 dx + ∫ 1 δ1 ex + x √ex − 1 dx Note that the second integral is

3 3. (25 points) (a) (19 points) i. (3 points) Let f : R → R be a function that is twice differentiable at x0 = 0. Define the terms: the Taylor polynomial and the Taylor remainder of order 2 of f about 0. Solution: The Taylor polynomial of order 2 of f about 0 is defined as T2(x) = f (0) + f ′(0)x + f ′′(0) 2 x2. The Taylor remainder of order 2 of f about 0 is defined as R2(x) = f (x) − T2(x) for every x in a neighbourhood of 0. ii. (4 points) Consider the function defined by f (x) = √1 + x for every x > −1. Compute the Taylor polynomial of f of order 2 about x0 = 0. Solution: We have f (x) = √1 + x ⇒ f (0) = 1; f ′(x) = 1 2 √1+x = 1 2 (1+x)− 1 2 ⇒ f ′(0) = 1 2 ; f ′′(x) = − 1 4 (1 + x)− 3 2 ⇒ f ′′(0) = − 1 4 . It follows that T2(x) = f (0) + f ′(0)x + f ′′(0) 2 x2 = 1 + x 2 − x2 8 iii. (1

4 (a) (19 points) Let (an)∞ n=1 , (bn)∞ n=1 be two sequences of real numbers. Suppose that an, bn ∈ N for every n ∈ N. Suppose also that the sequence (bn)∞ n=1 is strictly increasing. i. (4 points) Prove that limn→∞ bn = ∞. Solution: We prove by induction that bn ≥ n for every n ∈ N, implying that limn→∞ bn = ∞ by Toast theorem. Clearly, as b1 ∈ N then b1 ≥ 1. Now, suppose that bn ≥ n. As bn+1 > bn, and bn, bn+1 ∈ N, then bn+1 ≥ bn + 1 ≥ n + 1. ii. (15 points) Suppose, in addition, that for every 2 ≤ n ∈ N we have an · bn−1 − an−1 · bn = (−1)n. Prove that limn→∞ an bn exists. Solution: For every 2 ≤ n ∈ N we have an bn − an−1 bn−1 = an · bn−1 − an−1bn bnbn−1 = (−1)n bnbn−1 Thus, being a Telescopic sum, we have aN bN − a1 b1 = N ∑ n=2 ( an bn − an−1 bn−1 ) = N ∑ n=2 (−1)n bnbn−1 Note that a

5 Note that ∞ ∑ n=1 xn (n − 1)! = x ∞ ∑ n=1 xn−1 (n − 1)! = x ∞ ∑ n=0 xn n! = xex It follows that f (x) = x · (xex)′ = x(ex + xex) = x2ex + xex for every x ∈ R. Finally, f (1) = ∑∞ n=0 n2 n! = 2e. (b) (9 points) Let a, b ∈ R such that a < b and let f be a function that is defined on [a, b]. Suppose that f 2 is integrable on [a, b] and that ∫ b a f 2(x) dx = 0. Let x0 ∈ (a, b). Prove that if f is continuous at x0, then f (x0) = 0. Solution: Consider the function defined by F(x) = ∫ x a f 2(t) dt, for every a ≤ x ≤ b. Note that 0 ≤ F(x) ≤ ∫ b a f 2(x) dx for every a ≤ x ≤ b, implying that F(x) = 0 for every x ∈ [a, b]. As f is continuous at x0, then f 2 is continuous at x0. Thus, by the Fundamental theorem of calculus, f 2(x0) = F′(x0) = 0, which implies that f (x0) = 0. BEHATZLAHA!!!!

Infinitesimal Calculus 2, Moed A 2023, Reichman university. Do not write at the back of the papers!! Lecturer: Dr. Yossi Shamai Instructions: 1. Duration: 3 hours. Instructions: The following exam has 5 questions. You must answer exactly 4 questions. Each question is worth 25 credits. You must explain in full details all of your answers in all parts of the exam, and state all of the conditions for every theorem and/or claim. Unexplained answers will not receive any credit. You do not have a choice between different items of the same question. It is forbidden to rely on any material that was not learned in the course. You must write your solutions only in the designated areas. Do not write at the back of the papers!

2 1. (25 points) (a) (15 points) Prove that e is an irrational number. You may rely on any of the previous theorems or claims that were learned in the course. Solution: See lectures. (b) (10 points) Compute the value of the following definite integral ∫ π2 0 cos ( √x) dx Solution: Substituting t = √x then dt = 1 2 √x dx and hence dx = 2 √xdt = 2tdt. Thus ∫ π2 0 cos ( √x) dx = 2 ∫ π 0 t cos(t) dt = 2 [ [t sin t]π 0 − ∫ π 0 sin t dt ] = −2 ∫ π 0 sin t dt = 2 [cos t]π 0 = −4 2. (25 points) (a) (17 points) i. (2 points) Let x0 ∈ R and let f be a function that is differentiable ∞ times at x0. Define the term: the Taylor series of f about x0. Solution: The Taylor series of f about x0 is defined as T f (x) = ∞ ∑ n=0 f (n)(x0) n! · (x − x0)n ii. (10 points) Find the Taylor series about x0 = 0 of t

3 that f (c) = F′(c) = 0 3. (25 points) (a) (10 points) Let (an)∞ n=1 and (bn)∞ n=1 be two sequences of real numbers. Suppose that ∑∞ n=1 an converges conditionally, and that ∑∞ n=1 bn converges absolutely. Prove that ∑∞ n=1(an − bn) converges conditionally. Solution: First, as ∑∞ n=1 bn converges absolutely then it converges, which implies that by the Algebra of series, ∑∞ n=1(an − bn) is convergent, as it is a difference of two convergent series. To prove that it converges conditionally, we must prove that it does not converge absolutely. ATC that ∑∞ n=1 |an − bn| < ∞. Then by the comparison principle for nonnegative series, ∞ ∑ n=1 |an| = ∞ ∑ n=1 |an − bn + bn| ≤ ∞ ∑ n=1 (|an − bn| + |bn|) = ∞ ∑ n=1 |an − bn| + ∞ ∑ n=1 |bn| < ∞ in contradiction to the assumption that ∑∞ n=1 an converges

4 and let G be an anti-derivative of f in [a, b]. Prove that ∫ b a f (x) dx = G(b) − G(a) Solution: See lectures. (b) (16 points) i. (8 points) Prove that for every a > 0 there exists a k ∈ N such that for every n ≥ k we have n! ≥ an. Solution: Let a > 0. We have seen that the power series ∑∞ n=0 xn n! has a convergence radius R = ∞. Thus, the series converges at the point x = a and by the necessary criterion, we obtain that limn→∞ an n! = 0. It follows that there exists a k ∈ N for which an n! ≤ 1 for every n ≥ k. Multiplying by n! both sides of the inequality yields that n! ≥ an for every n ≥ k. ii. (8 points) Let (an)∞ n=0 be a sequence of positive numbers. Let R be the convergence radius of the power series ∑∞ n=0 an xn and suppose that R < ∞. Prove that the series ∑∞ n=0 ann!xn diverg

5 (b) (10 points) Let F : R2 → R be the function that is defined by F(x, y) =    xy(x2−y2) x2+y2 (x, y) , (0, 0) 0 (x, y) = (0, 0) i. (3 points) Compute the value of F′ x(0, 0). ii. (3 points) Compute the value of F′ x(2, 2). iii. (4 points) Compute the value of F′′ xy(0, 0). Solution: (i) We have F′ x(0, 0) = lim x→0 F(x, 0) − F(0, 0) x = lim x→0 0 − 0 x = 0 (ii) For every (x, y) , (0, 0) we have F′ x(x, y) = (x2 + y2)(y(x2 − y2) + 2x2y) − 2x2y(x2 − y2) (x2 + y2)2 Thus, Fx(2, 2) = 2. (iii) We compute F′′ xy(0, 0) = lim y→0 F′ x(0, y) − F′ x(0, 0) y = lim y→0 − y5 y4 − 0 y = lim y→0(−1) = −1

Infinitesimal Calculus 2, Moed A 2024, Reichman university. Do not write at the back of the papers!! Lecturer: Dr. Yossi Shamai Instructions: 1. Duration: 3 hours. Instructions: The following exam has 5 questions. You must answer exactly 4 questions. Each question is worth 25 credits. You must explain in full details all of your answers in all parts of the exam, and state all of the conditions for every theorem and/or claim. Unexplained answers will not receive any credit. You do not have a choice between different items of the same question. It is forbidden to rely on any material that was not learned in the course. You must write your solutions only in the designated areas. Do not write at the back of the papers! Behatzlacha!!!

2 1. (25 points) 1.1. (15 points) Let (an)∞ n=1 be a sequences of real numbers. Suppose that the series ∑∞ n=1(−1)n+1an is a Leibniz series. Prove that it converges to a value s ∈ R, and that a1 − a2 ≤ s ≤ a1. Solution: See lectures. 1.2. (10 points) Compute the following definite integral: ∫ 1 0 arccos(x) dx Solution: We use integration by parts, ∫ 1 0 arccos(x) dx u′=1, v=arccos(x) = [x · arccos(x)]1 0 + ∫ 1 0 x √1 − x2 dx =t=1−x2⇒dt=−2xdx = 0 − 1 2 ∫ 0 1 dt √t = 1 2 ∫ 1 0 dt √t dt formula = 1 2 · 1 1 − 1 2 = 1 We now give an alternative solution: we use the method of substitution: As sin(t) ≥ 0 for every t ∈ [0, π 2 ] and as cos2(t) + sin2(t) = 1 then sin(t) = √1 − cos2(t) for every t ∈ [0, π 2 ]. We now compute ∫ 1 0 arccos(x) dx t=arccos(x)→dt=− dx √1−x2 = − ∫ 0 π 2 t √1 − x2 dt = ∫ π

3 Solution: False. ATC that the series is convergent. Applying parentheses on every 3 terms, then the resulting series is convergent by the parentheses test. But (1 − 1 + 1) + ( 1 2 − 1 2 + 1 2 ) + ( 1 3 − 1 3 + 1 3 ) + · · · = 1 + 1 2 + 1 3 + · · · = ∞ ∑ n=1 1 n = ∞ Contradiction. 3. (25 points) 3.1. (17 points) Let f be a function that is defined on [0, 1]. 3.1.1. (2 points) Define the term: f is differentiable on [0, 1). Note: you do not have to use the − δ language. Solution: We say that f is differentiable on [0, 1) if for every x0 ∈ (0, 1), the limit limx→x0 f (x)− f (x0) x−x0 exists, and also the right-sided limit limx→0+ f (x)− f (0) x exists. 3.1.2. (15 points) Suppose that f is continuous on [0, 1] and differentiable on [0, 1). Suppose also that f (0) = 0 and that f ′(0) = f (1

4 4.1.1. (6 points) Compute the minimal value of f or prove that it doesn’t exist. Solution: Note that f is differentiable by AOD and for every x > 0 we have f ′(x) = 1 2 ( 1 − 2 x2 ) = 1 2 x2 − 2 x2 = 0 ⇔ x = √2 As f ′(x) ≥ 0 for every x ≥ √2 and f ′(x) ≤ 0 for every 0 < x ≤ √2 then f is decreasing on (0, √2] and increasing on [ √2, ∞). Thus, x = √2 is a minimum of f , and the minimal value is f ( √2) = √2. 4.1.2. (7 points) Prove that the sequence (an)∞ n=1 is well defined, and that it is convergent. Solution: As a1 = 2 > √2 and for every n ∈ N we have an+1 = f (an) ≥ √2 then an > 0 for all n ∈ N, and therefore the sequence (an)∞ n=1 is well defined and bounded from below by √2. We now prove by induction that the sequence (an)∞ n=1 is decreasing. First, we have a2 = 1.5 ≤ 2 = a1. Let n ∈

5 Solution: Consider the power series ∑∞ n=0 (−1)n(n+1)! n! xn = ∑∞ n=0(−1)n(n + 1)xn. Note that the convergence radius is given by R = 1 L = lim n→∞ ∣∣∣∣∣ an an+1 ∣∣∣∣∣ = lim n→∞ n + 1 n + 2 = lim n→∞ 1 + 1 n 1 + 2 n = 1 Thus the series converges on the open interval (−1, 1). Consider the function f (x) = ∑∞ n=0 (−1)n(n+1)! n! xn for every x ∈ (−1, 1). Then f is well defined and therefore, by the uniqueness theorem, for every n ≥ 0 we have f (n)(0) n! = an ⇒ f (n)(0) = n! · an = (−1)n(n + 1)! 5. (25 points) 5.1. (8 points) Prove that ex ≥ 1 + x + x2 2 for every x ≥ 0. Solution: Consider the function defined by f (x) = ex − 1 − x − x2 2 for every x ≥ 0. Note that f is differentiable by AOD and that f ′(x) = ex − 1 − x for every x ≥ 0. As we have proved in class that ex ≥ 1 + x, ∀x ≥ 0, the

Calculus 2, Moed A 2025 Reichman university Solutions Lecturer: Dr. Yossi Shamai. Duration: 3 hours. Auxiliary material: none. Instructions: The following exam has 5 questions. You must answer exactly 4 questions. Each question is worth 25 credits. There is no choice between different items of the same question. Unexplained answers will not receive any credit. You must state all of the conditions for every theorem and/or claim. You do not have a choice between different items of the same question. It is forbidden to rely on any material that was not learned in the course. You must write your solutions only in the designated areas. Do not write at the back of the papers! Good Luck! 1. (25 points) 1.1. (17 points) Let a, b ∈ R such that a < b and let f, G be two functions that are defined on