סימולציה 4 — מבחן מעורב

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 (1 + 𝑥) − 𝑥 0 0 = lim 𝑥→0 1 cos2 ( 𝑥) − 1 1 1+𝑥 − 1 = lim 𝑥→0 h 1−cos2 ( 𝑥) cos2 ( 𝑥) i h 1− (1+𝑥) 1+𝑥 i formula = lim 𝑥→0 sin2 (𝑥) · (1 + 𝑥) cos2 (𝑥) · (−𝑥) = lim 𝑥→0 − sin(𝑥) 𝑥 · sin(𝑥) · 1 cos2 (𝑥) · (1 + 𝑥) AOL = 0 1

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, in addition, that 𝑓 is continuous at 𝑥0. Prove that 𝑓 is continuously differentiable at 𝑥0. Solution: As 𝑓 is continuous at 𝑥0, then lim 𝑥→𝑥0 𝑓 (𝑥) = 𝑓 (𝑥0). It follows that lim 𝑥→𝑥0 [ 𝑓 (𝑥) − 𝑓 (𝑥0)] AOL = 0 Thus, by L’Hˆopital’s rule 𝑓 ′ (𝑥0) = lim 𝑥→𝑥0 𝑓 (𝑥) − 𝑓 (𝑥0) 𝑥 − 𝑥0 0 0 = lim 𝑥→𝑥0 𝑓 ′ (𝑥) 1 = lim 𝑥→𝑥0 𝑓 ′ (𝑥) = 𝐿 It follows that 𝑓 is continuously

Calculus II – Spring 2025-26 Homework 6 Solution 1. Compute the value of each of the following series, or prove that it diverges. (a) ∞ ∑ 𝑛=0 (−1)𝑛 · 22𝑛+1 + (−1)𝑛 5𝑛 Solution: We have ∞ ∑ 𝑛=0 (−1)𝑛 · 22𝑛+1 + (−1)𝑛 5𝑛 = ∞ ∑ 𝑛=0 (−1)𝑛 · 2 · 4𝑛 5𝑛 + ∞ ∑ 𝑛=0 (−1)𝑛 · (−1)𝑛 5𝑛 = 2 · ∞ ∑ 𝑛=0 ( − 4 5 )𝑛 + ∞ ∑ 𝑛=0 ( 1 5 )𝑛 = 2 · 1 1 + 4 5 + 1 1 − 1 5 = 10 9 + 5 4 = 85 36 (b) ∞ ∑ 𝑛=1 cosh ( 1 𝑛 ) , where cosh(𝑥) = 𝑒𝑥 + 𝑒−𝑥 2 , ∀𝑥 ∈ R Solution: Note that lim 𝑛→∞ 𝑎𝑛 = lim 𝑛→∞ cosh ( 1 𝑛 ) Heine’s lemma = lim 𝑥→0 cosh(𝑥) = lim 𝑥→0 𝑒𝑥 + 𝑒−𝑥 2 = 1 Thus, the series diverges by the necessary criterion. (c) ∞ ∑ 𝑛=1 𝑛2 − 1 𝑛3 Solution: Note that the series is non-negative. We have 𝐿 = lim 𝑛→∞ 𝑎𝑛 𝑏𝑛 = lim 𝑛→∞ [ 𝑛2 −1 𝑛3 ] [ 1 𝑛 ] = lim 𝑛→∞ 𝑛3 − 𝑛 𝑛3 = lim 𝑛→∞ [ 1 − 1 𝑛2 ] = 1 It follows that the series is divergent together with the Harmonic series. 1

Alternative solution 1: We have ∞ ∑ 𝑛=1 𝑛2 − 1 𝑛3 ≥ ∞ ∑ 𝑛=1 𝑛2 − 𝑛 𝑛3 = ∞ ∑ 𝑛=1 𝑛 − 1 𝑛2 = ∞ ∑ 𝑛=0 𝑛 (𝑛 + 1)2 = ∞ ∑ 𝑛=1 𝑛 (𝑛 + 1)2 ≥ ∞ ∑ 𝑛=1 𝑛 (𝑛 + 𝑛)2 = ∞ ∑ 𝑛=1 𝑛 4𝑛2 = 1 4 ∞ ∑ 𝑛=1 1 𝑛 = ∞ By the comparison test, ∑∞ 𝑛=1 𝑛2 −1 𝑛3 = ∞. Alternative solution 2: We have ∞ ∑ 𝑛=1 𝑛2 − 1 𝑛3 = ∞ ∑ 𝑛=1 ( 1 𝑛 − 1 𝑛3 ) = divergent − convergent = divergent (d) ∞ ∑ 𝑛=1 ln ( 2𝑛 + 3 2𝑛 + 1 ) Solution: We have ∞ ∑ 𝑛=1 ln ( 2𝑛 + 3 2𝑛 + 1 ) ln ( 𝑥 𝑦 ) =ln 𝑥−ln 𝑦 = ∞ ∑ 𝑛=1 [ln(2𝑛 + 3) − ln(2𝑛 + 1)] definition = lim 𝑁 →∞ 𝑁 ∑ 𝑛=1 ln ( 2𝑛 + 3 2𝑛 + 1 ) = lim 𝑁 →∞ 𝑁 ∑ 𝑛=1 [ln(2𝑛 + 3) − ln(2𝑛 + 1)] = lim 𝑁 →∞ [HHH ln(5) − ln(3) + HHH ln(7) − HHH ln(5) + · · · + ln(2𝑁 + 3) − XXXXX ln(2𝑁 + 1)] = lim 𝑁 →∞ [ln(2𝑁 + 3) − ln(3)] Heine’s lemma = lim 𝑥→∞ [ln 𝑥 − ln 3] = ∞ Hence the series is divergent. Alternative solution: Note that for every 𝑛 ∈ N we have 2𝑛+3 2𝑛+1 ≥ 1 which implies that ln ( 2𝑛+3 2𝑛+1 ) ≥ 0. It follows that the series is non-negative. We now compute ∞ ∑ 𝑛=1 ln ( 2𝑛 + 3 2𝑛 + 1 ) = ∞ ∑ 𝑛=1 ln ( 1 + 2 2𝑛 + 1 ) ln(1+𝑥) ≥ 𝑥 𝑥+1 ≥ ∞ ∑ 𝑛=1 ( 2 2𝑛+1 ) (1 + 2 2𝑛+1 ) = ∞ ∑ 𝑛=1 2 2𝑛 + 3 ≥ ∞ ∑ 𝑛=1 2 2𝑛 + 3𝑛 = 2 5 ∞ ∑ 𝑛=1