The Lumenary · Early undergraduate

Calculus of one variable

Name: ______________________

Date: ______________________

You will learn to

Use the ε–δ definitions of limit and continuity, and prove continuity of simple functions directly from them
Differentiate and integrate elementary functions with the chain rule, substitution and integration by parts, and justify these rules
State and prove Rolle's theorem, the mean value theorem and L'Hôpital's rule, and use them to analyse functions and limits
Prove the fundamental theorem of calculus from the mean value theorem, and evaluate proper and improper integrals
Test series for convergence (comparison, ratio, integral and alternating series tests) and prove that absolute convergence implies convergence
Compute Taylor polynomials with the Lagrange error bound, prove $e^x = \sum x^n/n!$ , and explain why a smooth function need not equal its Taylor series

1.
We want $|x^2 - 9| < \varepsilon$ whenever $|x - 3| < \delta$ . Following the lesson's recipe $\delta = \min\big(1, \varepsilon/(2|a| + 1)\big)$ with $a = 3$ , which $\delta$ does it give for $\varepsilon = 0.7$ ?
- A. $\delta = 0.1$
- B. $\delta = 0.7$
- C. $\delta = 1$
- D. $\delta = 0.7/6$
2.
Compute $\displaystyle\lim_{x \to 0} \frac{1 - \cos(4x)}{x^2}$ .
3.
For $f(x) = |x|$ on $[-1, 2]$ , the chord slope is $\frac{f(2) - f(-1)}{2 - (-1)} = \frac13$ , but $f'(x) \in \{-1, 1\}$ wherever it exists. Which hypothesis of the mean value theorem fails?
- A.Differentiability on the open interval $(-1, 2)$
- B.Continuity on $[-1, 2]$
- C.The condition $f(a) = f(b)$
- D.None — the theorem is false
4.
Compute $\displaystyle\int_1^e \ln x\,dx$ .
5.
Which of these series converge? (Select all that apply.)
- A. $\displaystyle\sum_{n \ge 1} \frac{n!}{n^n}$
- B. $\displaystyle\sum_{n \ge 1} \frac{(-1)^{n+1}}{\sqrt n}$
- C. $\displaystyle\sum_{n \ge 2} \frac{1}{n \ln n}$
- D. $\displaystyle\sum_{n \ge 1} \frac{1}{n^{1.01}}$
- E. $\displaystyle\sum_{n \ge 1} \frac{n}{n + 1}$
6.
Using the Lagrange bound $|e - T_n(1)| \le \dfrac{e}{(n + 1)!}$ for the Maclaurin polynomial of $e^x$ at $x = 1$ , what is the smallest degree $n$ that guarantees $|e - T_n(1)| < 10^{-6}$ ? (Use $e < 3$ if you like; the answer is the same.)
7.
A student writes: $\displaystyle\lim_{x \to 0} \frac{x^2 + 1}{x + 1} = \lim_{x \to 0} \frac{2x}{1} = 0$ . What is wrong?
- A.L'Hôpital's rule does not apply: the quotient is not an indeterminate form, and the limit is $1$ .
- B.Nothing — the limit is $0$ .
- C.The derivative of $x^2 + 1$ is wrong.
- D.One should differentiate the quotient with the quotient rule instead.
8.
Let $f(x) = e^{-1/x^2}$ for $x \ne 0$ and $f(0) = 0$ . On which set does $f$ equal the sum of its Maclaurin series?
- A.Only at $x = 0$
- B.On all of $\R$ , because $f$ is smooth
- C.On $(-1, 1)$ , the radius of convergence
- D.Nowhere, because the series diverges
9.
Which Lean statement correctly formalises the mean value theorem? (Here Icc a b is $[a, b]$ and Ioo a b is $(a, b)$ .)
- A.(hab : a < b) (hfc : ContinuousOn f (Icc a b)) (hff' : ∀ x ∈ Ioo a b, HasDerivAt f (f' x) x) : ∃ c ∈ Ioo a b, f' c = (f b - f a) / (b - a)
- B.(hfc : ContinuousOn f (Icc a b)) (hff' : ∀ x ∈ Ioo a b, HasDerivAt f (f' x) x) : ∃ c ∈ Ioo a b, f' c = (f b - f a) / (b - a)
- C.(hab : a < b) (hfc : ContinuousOn f (Icc a b)) (hff' : ∀ x ∈ Ioo a b, HasDerivAt f (f' x) x) : ∀ c ∈ Ioo a b, f' c = (f b - f a) / (b - a)
- D.(hab : a < b) (hff' : ∀ x ∈ Ioo a b, HasDerivAt f (f' x) x) : ∃ c ∈ Ioo a b, f' c = (f b - f a) / (b - a)
10.
Write exp_taylor(x, n) returning $T_n(x) = \sum_{k=0}^{n} x^k/k!$ , and degree_needed(x, tol) returning the least $n \ge 0$ for which the Lagrange bound $e^{\max(x, 0)}\,|x|^{n+1}/(n+1)!$ is less than tol. Then check that $T_n(x)$ really is within tol of math.exp(x).
Computational — Practice

Answer key — Calculus of one variable

1.
With $a = 3$ , $2|a| + 1 = 7$ , so $\delta = \min(1, 0.7/7) = 0.1$ . Then $|x - 3| < 0.1$ gives $|x + 3| \le |x - 3| + 6 < 6.1 < 7$ and $|x^2 - 9| = |x - 3|\,|x + 3| < 0.1 \cdot 7 = 0.7$ .
2.
By L'Hôpital twice: $\lim \frac{1 - \cos 4x}{x^2} = \lim \frac{4 \sin 4x}{2x} = \lim \frac{16\cos 4x}{2} = 8$ . Each step is legitimate because the previous quotient is $0/0$ and the next limit exists. With Taylor's theorem instead: $1 - \cos(4x) = 8x^2 + O(x^4)$ .
3.
$f(x) = |x|$ is not differentiable at $0$ , which lies in $(-1, 2)$ . Since $f'$ is only ever $\pm 1$ , no point has $f'(c) = 1/3$ : the hypothesis is genuinely necessary.
4.
With $u = \ln x$ , $dv = dx$ : $\int \ln x\,dx = x\ln x - \int x \cdot \frac1x\,dx = x\ln x - x$ . By the fundamental theorem, $\int_1^e \ln x\,dx = (e - e) - (0 - 1) = 1$ .
5.
(a) converges by the ratio test (limit $1/e$ ); (b) converges conditionally by the alternating series test; (c) diverges by the integral test ( $\ln \ln b \to \infty$ ); (d) converges ( $p > 1$ ); (e) diverges because $\frac{n}{n+1} \to 1 \ne 0$ .
6.
We need $(n+1)! > e \cdot 10^{6} \approx 2.72 \times 10^6$ . Since $9! = 362\,880$ is too small and $10! = 3\,628\,800$ is large enough, $n + 1 = 10$ , so $n = 9$ . (The actual error of $T_9(1)$ is about $3.0 \times 10^{-7}$ ; the bound is honest, if a little pessimistic.)
7.
The numerator tends to $1$ and the denominator to $1$ , so this is not a $0/0$ form and L'Hôpital's rule does not apply. The quotient is continuous at $0$ , so the limit is $\frac{0 + 1}{0 + 1} = 1$ .
8.
The lesson proves $f^{(n)}(0) = 0$ for all $n$ , so the Maclaurin series is the zero series. It converges everywhere, but to $0$ , and $f(x) = e^{-1/x^2} > 0$ for $x \ne 0$ . So $f$ equals its Maclaurin series only at $0$ : $f$ is smooth but not analytic at $0$ .
9.
Option (a). The theorem needs $a < b$ (otherwise the open interval is empty), continuity on $[a, b]$ (otherwise redefining $f(b)$ breaks it), differentiability on $(a, b)$ , and concludes that some $c \in (a, b)$ has $f'(c)$ equal to the chord slope.

10.

import math

def exp_taylor(x, n):
    term, total = 1.0, 1.0
    for k in range(1, n + 1):
        term *= x / k
        total += term
    return total

def degree_needed(x, tol):
    n = 0
    while math.exp(max(x, 0)) * abs(x) ** (n + 1) / math.factorial(n + 1) >= tol:
        n += 1
    return n

For $x = 1$ and $10^{-6}$ this gives $n = 9$ , as in the earlier exercise. The tests confirm the guarantee numerically for a few $x$ ; Taylor's theorem is what makes it hold for every $x$ .