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סימולציה 4 — מבחן מעורב

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

מעורב120 דקות

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

Calculus II – Spring2025−26Homework1SolutionSpring 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→0x2−xsin⁡x(ex−1)⋅ln⁡(1+x)\lim_{x \to 0} \frac{x^2 - x \sin x}{(e^x - 1) \cdot \ln(1+x)}

Solution: First note that

lim⁡x→0xex−1=00=lim⁡x→01ex=1(★)\lim_{x \to 0} \frac{x}{e^x - 1} = \frac{0}{0} = \lim_{x \to 0} \frac{1}{e^x} = 1 \quad \text{(★)}

and

lim⁡x→0x−sin⁡xln⁡(1+x)=00=lim⁡x→01−cos⁡x11+x=0(★★)\lim_{x \to 0} \frac{x - \sin x}{\ln(1 + x)} = \frac{0}{0} = \lim_{x \to 0} \frac{1 - \cos x}{\frac{1}{1+x}} = 0 \quad \text{(★★)}

It follows that

lim⁡x→0x2−xsin⁡x(ex−1)⋅ln⁡(1+x)=lim⁡x→0x(x−sin⁡x)(ex−1)⋅ln⁡(1+x)=lim⁡x→0[xex−1⋅x−sin⁡xln⁡(1+x)]\lim_{x \to 0} \frac{x^2 - x \sin x}{(e^x - 1) \cdot \ln(1 + x)} = \lim_{x \to 0} \frac{x(x - \sin x)}{(e^x - 1) \cdot \ln(1 + x)} = \lim_{x \to 0} \left[\frac{x}{e^x - 1} \cdot \frac{x - \sin x}{\ln(1 + x)}\right]
=(★), (★★)0\stackrel{\text{(★), (★★)}}{=} 0

(b) lim⁡x→0+x2⋅ln⁡(x)\lim_{x \to 0^+} x^2 \cdot \ln(x)

Solution: We have

lim⁡x→0+x2⋅ln⁡(x)=lim⁡x→0+ln⁡(x)1x2=−∞∞=lim⁡x→0+1x−2x3=lim⁡x→0+[−x22]\lim_{x \to 0^+} x^2 \cdot \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x^2}} = \frac{-\infty}{\infty} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \to 0^+} \left[-\frac{x^2}{2}\right]
=L’H0\stackrel{\text{L'H}}{=} 0

(c) lim⁡x→0tan⁡(x)−xln⁡(1+x)−x\lim_{x \to 0} \frac{\tan(x) - x}{\ln(1+x) - x}

Solution: We have

lim⁡x→0tan⁡(x)−xln⁡(1+x)−x=00=lim⁡x→01cos⁡2(x)−111+x−1\lim_{x \to 0} \frac{\tan(x) - x}{\ln(1+x) - x} = \frac{0}{0} = \lim_{x \to 0} \frac{\frac{1}{\cos^2(x)} - 1}{\frac{1}{1+x} - 1}
=lim⁡x→0[1−cos⁡2(x)cos⁡2(x)][1−(1+x)1+x]=lim⁡x→0sin⁡2(x)⋅(1+x)cos⁡2(x)⋅(−x)= \lim_{x \to 0} \frac{\left[\frac{1 - \cos^2(x)}{\cos^2(x)}\right]}{\left[\frac{1 - (1+x)}{1+x}\right]} = \lim_{x \to 0} \frac{\sin^2(x) \cdot (1 + x)}{\cos^2(x) \cdot (-x)}
=lim⁡x→0[−sin⁡(x)x⋅sin⁡(x)⋅1cos⁡2(x)⋅(1+x)]= \lim_{x \to 0} \left[-\frac{\sin(x)}{x} \cdot \sin(x) \cdot \frac{1}{\cos^2(x)} \cdot (1 + x)\right]
=L’H0\stackrel{\text{L'H}}{=} 0
2
קשהבינונימטלה
גבולותטוריםנגזרות
2.Let x0∈Rx_0 \in \mathbb{R} and let ff be a function that is defined on a neighborhood of x0x_0. Suppose that ff is differentiable at every xx in the neighborhood for which x≠x0x \neq x_0. Let L∈RL \in \mathbb{R} and suppose also that lim⁡x→x0f′(x)=L\lim_{x \to x_0} f'(x) = L.

(a) Give an example of a function satisfying the conditions above, and such that ff is not differentiable at x0x_0.

Solution:

Consider the function defined by

f(x)={0x≠01x=0f(x) = \begin{cases} 0 & x \neq 0 \\ 1 & x = 0 \end{cases}

for every x∈Rx \in \mathbb{R}. We choose x0=0x_0 = 0. Note that for every x≠0x \neq 0, ff is differentiable by AOD at xx and that for every x≠0x \neq 0 we have f′(x)=0f'(x) = 0. It follows that

lim⁡x→0f′(x)=lim⁡x→00=0\lim_{x \to 0} f'(x) = \lim_{x \to 0} 0 = 0

Note also that

lim⁡x→0f(x)=lim⁡x→00=0≠1=f(0),\lim_{x \to 0} f(x) = \lim_{x \to 0} 0 = 0 \neq 1 = f(0),

which implies that ff is not continuous at x0=0x_0 = 0 and therefore ff is not differentiable at x0=0x_0 = 0.

(b) Suppose, in addition, that ff is continuous at x0x_0. Prove that ff is continuously differentiable at x0x_0.

Solution:

As ff is continuous at x0x_0, then lim⁡x→x0f(x)=f(x0)\lim_{x \to x_0} f(x) = f(x_0). It follows that

lim⁡x→x0[f(x)−f(x0)]=0\lim_{x \to x_0} [f(x) - f(x_0)] = 0

Thus, by L'Hôpital's rule

f′(x0)=lim⁡x→x0f(x)−f(x0)x−x0=lim⁡x→x0f′(x)1=lim⁡x→x0f′(x)=Lf'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0} \frac{f'(x)}{1} = \lim_{x \to x_0} f'(x) = L

It follows that ff is continuously differentiable at x0x_0.

3
קשהבינונימטלה
גבולותטוריםמבחן-ההשוואה

# Calculus II – Spring2025−26Homework6SolutionSpring 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)n5n\sum_{n=0}^{\infty} (-1)^n \cdot \frac{2^{2n+1} + (-1)^n}{5^n}

Solution: We have

∑n=0∞(−1)n⋅22n+1+(−1)n5n=∑n=0∞(−1)n⋅2⋅4n5n+∑n=0∞(−1)n⋅(−1)n5n\sum_{n=0}^{\infty} (−1)^n · \frac{2^{2n+1} + (−1)^n}{5^n} = \sum_{n=0}^{\infty} (−1)^n · \frac{2 · 4^n}{5^n} + \sum_{n=0}^{\infty} \frac{(−1)^n · (−1)^n}{5^n}
=2⋅∑n=0∞(−45)n+∑n=0∞(15)n= 2 · \sum_{n=0}^{\infty} \left( − \frac{4}{5} \right)^n + \sum_{n=0}^{\infty} \left( \frac{1}{5} \right)^n
=2⋅11+45+11−15=109+54=8536= 2 · \frac{1}{1 + \frac{4}{5}} + \frac{1}{1 − \frac{1}{5}} = \frac{10}{9} + \frac{5}{4} = \frac{85}{36}

### (b) ∑n=1∞cosh⁡(1n)\sum_{n=1}^{\infty} \cosh \left( \frac{1}{n} \right), where cosh⁡(x)=ex+e−x2,∀x∈R\cosh(x) = \frac{e^x + e^{-x}}{2}, \forall x \in \mathbb{R}

Solution: Note that

lim⁡n→∞an=lim⁡n→∞cosh⁡(1n)=Heine’s lemmalim⁡x→0cosh⁡(x)=lim⁡x→0ex+e−x2=1\lim_{n \to \infty} a_n = \lim_{n \to \infty} \cosh \left( \frac{1}{n} \right) \stackrel{\text{Heine's lemma}}{=} \lim_{x \to 0} \cosh(x) = \lim_{x \to 0} \frac{e^x + e^{-x}}{2} = 1

Thus, the series diverges by the necessary criterion.

### (c) ∑n=1∞n2−1n3\sum_{n=1}^{\infty} \frac{n^2 - 1}{n^3}

Solution: Note that the series is non-negative. We have

L=lim⁡n→∞anbn=lim⁡n→∞[n2−1n3][1n]=lim⁡n→∞n3−nn3=lim⁡n→∞[1−1n2]=1L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\left[\frac{n^2 −1}{n^3}\right]}{\left[\frac{1}{n}\right]} = \lim_{n \to \infty} \frac{n^3 − n}{n^3} = \lim_{n \to \infty} \left[ 1 − \frac{1}{n^2} \right] = 1

It follows that the series is divergent together with the Harmonic series.

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בינוניבינונימטלה
גבולותטוריםמבחן-ההשוואה
2.Determine whether each of the following series converges or diverges.

(a) ∑n=2∞1ln⁡(n)\sum_{n=2}^{\infty} \frac{1}{\ln(n)}

Solution:

Note that 1ln⁡(n)≥0\frac{1}{\ln(n)} \geq 0 for every n≥3n \geq 3. We now write

∑n=3∞1ln⁡(n)≥∑n=3∞1n\sum_{n=3}^{\infty} \frac{1}{\ln(n)} \geq \sum_{n=3}^{\infty} \frac{1}{n}

tail of Harmonic =∞= \infty

It follows that the series ∑n=3∞1ln⁡(n)\sum_{n=3}^{\infty} \frac{1}{\ln(n)} is divergent by the comparison test, and therefore the series ∑n=2∞1ln⁡(n)\sum_{n=2}^{\infty} \frac{1}{\ln(n)} is divergent as it has a divergent tail.

(b) ∑n=1∞(n+2−n)3\sum_{n=1}^{\infty} \left(\sqrt{n + 2} - \sqrt{n}\right)^3

Solution:

We note that

∑n=1∞(n+2−n)3=∑n=1∞(n+2−nn+2+n)3=∑n=1∞8(n+2+n)3\sum_{n=1}^{\infty} \left(\sqrt{n + 2} - \sqrt{n}\right)^3 = \sum_{n=1}^{\infty} \left(\frac{n + 2 - n}{\sqrt{n + 2} + \sqrt{n}}\right)^3 = \sum_{n=1}^{\infty} \frac{8}{\left(\sqrt{n + 2} + \sqrt{n}\right)^3}

We have

lim⁡n→∞[8(n+2+n)3][1n3/2]=lim⁡n→∞8n3/2(n+2+n)3=lim⁡n→∞8(1+2n+1)3=1\lim_{n \to \infty} \frac{\left[\frac{8}{(\sqrt{n+2}+\sqrt{n})^3}\right]}{\left[\frac{1}{n^{3/2}}\right]} = \lim_{n \to \infty} \frac{8n^{3/2}}{(\sqrt{n + 2} + \sqrt{n})^3} = \lim_{n \to \infty} \frac{8}{\left(\sqrt{1 + \frac{2}{n}} + 1\right)^3} = 1

Thus, the series converges together with the 32\frac{3}{2}-Harmonic series.

(c) ∑n=1∞1(n+e)n−en\sum_{n=1}^{\infty} \frac{1}{(n + e)^n - e^n}

Solution:

By the comparison test

∑n=3∞1(n+e)n−en≤∑n=3∞1(e+e)n−en=∑n=3∞12n⋅en−en≤∑n=3∞12⋅en−en=∑n=3∞1en\sum_{n=3}^{\infty} \frac{1}{(n + e)^n - e^n} \leq \sum_{n=3}^{\infty} \frac{1}{(e + e)^n - e^n} = \sum_{n=3}^{\infty} \frac{1}{2^n \cdot e^n - e^n} \leq \sum_{n=3}^{\infty} \frac{1}{2 \cdot e^n - e^n} = \sum_{n=3}^{\infty} \frac{1}{e^n}

tail of geometric <∞< \infty

It follows that ∑n=3∞1(n+e)n−en<∞\sum_{n=3}^{\infty} \frac{1}{(n+e)^n - e^n} < \infty and therefore ∑n=1∞1(n+e)n−en<∞\sum_{n=1}^{\infty} \frac{1}{(n+e)^n - e^n} < \infty.