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סימולציה 4 — מבחן מעורב

4 שאלות · 120 דקות · הוכחות, טורים, גבולות, אינטגרלים

מעורב120 דקות

טיפ

טיפ: קבעי טיימר ל-120 דקות. נסי לפתור לבד לפני שאת מציצה ברמזים.
1
קשהבינונימטלה
גבולותטוריםנגזרות

Calculus II – Spring2025−26Homework1Solution1.ComputeeachofthefollowinglimitsorprovethatitdoesnSpring 2025-26 Homework 1 Solution 1. Compute each of the following limits or prove that it doesn’texistintheextendedsense(a)limt exist in the extended sense (a) lim 𝑥→0\to 0 𝑥2 −𝑥 sinsin 𝑥 (𝑒𝑥 −1)⋅ln(1+-1) \cdot ln(1+𝑥).Solution:Firstnotethatlim) . Solution: First note that lim 𝑥→0\to 0 𝑥 𝑒𝑥 −100=lim- 1 0 0 = lim 𝑥→01\to 0 1 𝑒𝑥 AOL=1(AOL = 1 (★)andlim) and lim 𝑥→0\to 0 𝑥 −sin- sin 𝑥 ln(1+ln (1 + 𝑥)00=lim) 0 0 = lim 𝑥→01−cos\to 0 1 - cos 𝑥 1 1+𝑥 AOL=0(AOL = 0 (★★)Itfollowsthatlim) It follows that lim 𝑥→0\to 0 𝑥2 − 𝑥 sinsin 𝑥 (𝑒𝑥 −1)⋅ln(1+- 1) \cdot ln (1 + 𝑥)=lim) = lim 𝑥→0\to 0 𝑥 (𝑥 −sin- sin 𝑥) (𝑒𝑥 −1)⋅ln(1+- 1) \cdot ln (1 + 𝑥)=lim) = lim 𝑥→0\to 0 𝑥 𝑒𝑥 − 1 · 𝑥 −sin- sin 𝑥 ln(1+ln (1 + 𝑥) (★), (★★)=0(b)lim) = 0 (b) lim 𝑥→0+\to 0+ 𝑥2⋅ln(2 \cdot ln(𝑥).Solution:Wehavelim) . Solution: We have lim 𝑥→0+\to 0+ 𝑥2⋅ln(2 \cdot ln(𝑥)=lim) = lim 𝑥→0+ln(\to 0+ ln(𝑥) 1 𝑥2?∞=lim2 ? \infty = lim 𝑥→0+1\to 0+ 1 𝑥 − 2 𝑥3=lim3 = lim 𝑥→0+h−\to 0+ h - 𝑥 2iAOL=0(c)lim2 i AOL = 0 (c) lim 𝑥→0tan(\to 0 tan( 𝑥) −𝑥 ln(1+ln(1+𝑥) −𝑥 . Solution:wehavelimSolution: we have lim 𝑥→0tan(\to 0 tan(𝑥) − 𝑥 ln(1+ln (1 + 𝑥) − 𝑥 00=lim0 0 = lim 𝑥→01cos2(\to 0 1 cos2 ( 𝑥) − 1 1 1+𝑥 −1=lim- 1 = lim 𝑥→0h1−cos2(\to 0 h 1-cos2 ( 𝑥) cos2 ( 𝑥) i h 1− (1+𝑥) 1+𝑥 iformula=limi formula = lim 𝑥→0sin2(\to 0 sin2 (𝑥) · (1 + 𝑥) cos2 (𝑥) · (−𝑥)=lim) = lim 𝑥→0−sin(\to 0 - sin(𝑥) 𝑥 · sin(sin(𝑥) · 1 cos2 (𝑥) · (1 + 𝑥)AOL=01) AOL = 0 1

2
קשהבינונימטלה
גבולותטוריםנגזרות
2.Let 𝑥0∈Randlet0 \in \mathbb{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 ∈Randsupposealsothatlim\in \mathbb{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 ≠01\ne 0 1 𝑥 =0forevery= 0 for every 𝑥 ∈R.Wechoose\in \mathbb{R}. We choose 𝑥0=0.Notethatforevery0 = 0. Note that for every 𝑥 ≠0,\ne 0, 𝑓 is differentiable by AOD at 𝑥 and that for every 𝑥 ≠0wehave\ne 0 we have 𝑓 ′ (𝑥)=0.Itfollowsthatlim) = 0. It follows that lim 𝑥→0\to 0 𝑓 ′ (𝑥)=lim) = lim 𝑥→00=0Notealsothatlim\to 0 0 = 0 Note also that lim 𝑥→0\to 0 𝑓 (𝑥)=lim) = lim 𝑥→00=0≠1=\to 00 = 0 \ne 1 = 𝑓 (0), which implies that 𝑓 is not continuous at 𝑥0=0andtherefore0 = 0 and therefore 𝑓 is not differentiable at 𝑥0=0.(b)Suppose,inaddition,that0 = 0. (b) Suppose, in addition, that 𝑓 is continuous at 𝑥0. Prove that 𝑓 is continuously differentiable at 𝑥0. Solution: As 𝑓 is continuous at 𝑥0,thenlim0, then lim 𝑥→𝑥0 𝑓 (𝑥) = 𝑓 (𝑥0).Itfollowsthatlim0). It follows that lim 𝑥→𝑥0 [ 𝑓 (𝑥) − 𝑓 (𝑥0)]AOL=0Thus,byL0)] AOL = 0 Thus, by L’Hˆopital’s rule 𝑓 ′ (𝑥0) =lim= lim 𝑥→𝑥0 𝑓 (𝑥) − 𝑓 (𝑥0) 𝑥 − 𝑥000=lim0 0 0 = lim 𝑥→𝑥0 𝑓 ′ (𝑥)1=lim) 1 = lim 𝑥→𝑥0 𝑓 ′ (𝑥) = 𝐿 It follows that 𝑓 is continuously
3
קשהבינונימטלה
גבולותטור-טיילורטורי-חזקות

Calculus II – Spring2025−26Homework10Solution1.Letfbethefunctiondefinedbyf(x)=∑∞n=1(−1)n+1x2n+12n(2n+1)forevery−1≤x≤1.(a)Provethatfiswell−defined.Solution:Weshowthattheseriesf(x)=∑∞n=1(−1)n+1x2n+12n(2n+1)convergenceforevery−1≤x≤1.Wenotethat∞∑n=1(−1)n+1x2n+12n(2n+1)=x∞∑n=1(−1)n+1x2n2n(2n+1)Wehave∞∑n=1(−1)n+1x2n2n(2n+1)=t=x2∞∑n=1(−1)n+1tn2n(2n+1)WecomputeR=1limn→∞n√Spring 2025-26 Homework 10 Solution 1. Let f be the function defined by f (x) = ∑\infty n=1 (-1)n+1x2n+1 2n(2n+1) for every -1 \le x \le 1. (a) Prove that f is well-defined. Solution: We show that the series f (x) = ∑\infty n=1 (-1)n+1x2n+1 2n(2n+1) convergence for every -1 \le x \le 1. We note that \infty ∑ n=1 (-1)n+1x2n+1 2n(2n + 1) = x \infty ∑ n=1 (-1)n+1x2n 2n(2n + 1) We have \infty ∑ n=1 (-1)n+1x2n 2n(2n + 1) = t=x2 \infty ∑ n=1 (-1)n+1tn 2n(2n + 1) We compute \mathbb{R} = 1 lim n\to \infty n √∣ ∣ ∣ (−1)n+12n(2n+1)(-1)n+1 2n(2n+1) ∣ ∣ ∣ =1Thustheseriesconvergesforevery= 1 Thus the series converges for every ∣ ∣x2∣ ∣ <1< 1 ⇒ −1<x<1.Wenotethatforx=1,−1wegetaLeibnizseries(orminusofaLeibnizseries),thustheseries∑∞n=1(−1)n+1⋅x2n+12n⋅(2n+1)convergesforevery−1≤x≤1.Thus,fiswell−defined.(b)Findanexplicitexpressionforfforevery−1≤x≤1.Solution:Weknowthatforevery−1<x≤1wehaveln(1+x)=∞∑n=1(−1)n+1xnnWesubstitutex=t2anddividebothsidesby2,yieldingln(1+t2)2=∞∑n=1(−1)n+1t2n2n1-1 < x < 1. We note that for x = 1, -1 we get a Leibniz series (or minus of a Leibniz series), thus the series ∑\infty n=1 (-1)n+1\cdot x2n+1 2n\cdot (2n+1) converges for every -1 \le x \le 1. Thus, f is well-defined. (b) Find an explicit expression for f for every -1 \le x \le 1. Solution: We know that for every -1 < x \le 1 we have ln(1 + x) = \infty ∑ n=1 (-1)n+1 xn n We substitute x = t2 and divide both sides by 2, yielding ln (1 + t2) 2 = \infty ∑ n=1 (-1)n+1 t2n 2n 1

4
בינוניבינונימטלה
גבולותטור-טיילורטורי-חזקות
2.Considertheseries∑∞n=0n+12n⋅(x−1)2n+1,wherexConsider the series ∑\infty n=0 n+1 2n\cdot (x-1)2n+1 , where x̸ =1.(a)Findthedomainofconvergenceoftheseries.Solution:Wenotethat∞∑n=0n+12n⋅(x−1)2n+1=1x−1∞∑n=0n+12n⋅(x−1)2nWeconsider∞∑n=0n+12n⋅(x−1)2n=t=1(x−1)2∞∑n=0n+12ntnWecomputeR=1limn→∞nn+12n=(= 1. (a) Find the domain of convergence of the series. Solution: We note that \infty ∑ n=0 n + 1 2n \cdot (x - 1)2n+1 = 1 x - 1 \infty ∑ n=0 n + 1 2n \cdot (x - 1)2n We consider \infty ∑ n=0 n + 1 2n \cdot (x - 1)2n = t= 1 (x-1)2 \infty ∑ n=0 n + 1 2n tn We compute \mathbb{R} = 1 lim n\to \infty n \sqrt{n+1}2n = (⋆) 2 (⋆)Wenotethatforeveryn∈Nwehave1≤nn+1≤nn+n=21nn1nAslimn→∞21n=1andlimn→∞n1n=1,thenbysqueezerulewegetthatlimn→∞nn+1=1.Thus,theseriesconvergesforevery) We note that for every n \in \mathbb{N} we have 1 \le n \sqrt{n}+ 1 \le n \sqrt{n}+ n = 2 1 n n 1 n As lim n\to \infty 2 1 n = 1 and lim n\to \infty n 1 n = 1, then by squeeze rule we get that lim n\to \infty n \sqrt{n}+ 1 = 1. Thus, the series converges for every ∣ ∣ ∣ ∣ ∣ 1 (x − 1)2 ∣ ∣ ∣ ∣ ∣ <2< 2 ⇔ x>1+12orx<1−12Wenotethatforx=1+12,1−12theseriesdiverges,sincewegettheseries∞∑n=0n+12n⋅(1x > 1 + 1 \sqrt{2}or x < 1 - 1 \sqrt{2}We note that for x = 1 + 1 \sqrt{2}, 1 - 1 \sqrt{2}the series diverges, since we get the series \infty ∑ n=0 n + 1 2n \cdot ( 1 ± 12−1)2n=∞∑n=0(n+1)andthenecessarycriteriondoesnothold.Thusthedomainofconvergenceisx>1+12orx<1−12.41 \sqrt{2}- 1 )2n = \infty ∑ n=0 (n + 1) and the necessary criterion does not hold. Thus the domain of convergence is x > 1 + 1 \sqrt{2}or x < 1 - 1 \sqrt{2}. 4