Recitation 9

Problem 1 Find $\lim_{x \to 0^+}\frac{x^x-1}{\ln (x+1)}$.

Solution The limit is of type $0/0$. Moreover, we have $\lim_{x\to 0}\frac{(x^x - 1)'}{(\ln(x+1))'} = \lim_{x\to 0}\frac{x^x(1 + \ln x)}{(1/(x+1))} = \lim_{x\to 0}(1+x)x^x(1 + \ln x) = -\infty.$ By l’Hôpital’s rule, the limit is $-\infty$.

Problem 2 Find $\lim_{x \to 0}(e^{x}+e^{-x}-\cos x)^{\frac{1}{x^2}}$.

Solution The limit is of type $1^\infty$. Let $f(x) = e^x + e^{-x}-\cos x - 1$. We have $\lim_{x \to 0}(e^{x}+e^{-x}-\cos x)^{\frac{1}{x^2}} = \lim_{x \to 0}\left[(1+f(x))^{\frac{1}{f(x)}}\right]^{\frac{f(x)}{x^2}}.$ Since $\lim_{x\to 0}f(x) = 0$, $\lim_{x \to 0}(1+f(x))^{\frac{1}{f(x)}} = e$. Moreover, $\lim_{x\to 0}f(x)/x^2$ is of type $0/0$, and $\lim_{x\to 0}\frac{f'(x)}{(x^2)'} = \lim_{x\to 0}\frac{e^x - e^{-x} + \sin x}{2x}$ is again of type $0/0$, and $\lim_{x\to 0}\frac{f''(x)}{(x^2)''} = \lim_{x\to 0}\frac{e^x + e^{-x} + \cos x}{2} = \frac{3}{2}.$ By l’Hôpital’s rule, $\lim_{x\to 0}\frac{f(x)}{x^2} = \frac{3}{2}$ and the limit is $e^{3/2}$.

Problem 3 $\lim_{x \to 0^+}\frac{e^{\sin (-1+\cos x)}\sin (-1+\cos x)}{(e^x-1)^2}$.

Solution If $\lim_{x \to 0}\frac{\sin (-1+\cos x)}{(e^x-1)^2}$ exists, then $\lim_{x \to 0}\frac{e^{\sin (-1+\cos x)}\sin (-1+\cos x)}{(e^x-1)^2} = \lim_{x\to 0}e^{\sin (-1+\cos x)}\lim_{x \rightarrow 0^+}\frac{\sin (-1+\cos x)}{(e^x-1)^2} = \lim_{x \to 0}\frac{\sin (-1+\cos x)}{(e^x-1)^2}.$ Note that $\lim_{x \to 0}\frac{\sin (-1+\cos x)}{(e^x-1)^2}$ is of type $0/0$ and

$\begin{aligned}\lim_{x \to 0}\frac{(\sin (-1+\cos x))'}{((e^x-1)^2)'} = \lim_{x\to 0}\frac{\cos(-1+\cos x)(-\sin x)}{2(e^x - 1)e^x} \\ = \lim_{x\to 0}\frac{-\cos(-1+\cos x)}{2e^x}\lim_{x\to 0}\frac{\sin x}{e^x - 1} = -\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{e^x - 1}\end{aligned}$ if $\lim_{x\to 0}\frac{\sin x}{e^x - 1}$ exists.
Finally, $\lim_{x\to 0}\frac{\sin x}{e^x - 1} = \lim_{x\to 0}\frac{\sin x}{x}\frac{x}{e^x - 1} = 1$. The answer is thus $\frac{-1}{2}$.

Problem 4 Use Taylor’s theorem to calculate the following limit: $\lim_{x \rightarrow 0}\frac{\cos x- 1 + \frac{x^2}{2}-\frac{x^4}{4!}}{x(\sin x - x+\frac{x^3}{3!})}$.

Solution The 6th order Taylor polynomial of $\cos x$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}.$ By Taylor’s theorem,

$\lim_{x\to 0}\frac{\cos x - (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!})}{x^6} = 0,$ and so $\lim_{x\to 0}\frac{\cos x - 1 + \frac{x^2}{2!} - \frac{x^4}{4!}}{x^6} = -\frac{1}{6!}.$ The 5th order Taylor polynomial of $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!}.$ By Taylor’s theorem, $\lim_{x\to 0}\frac{\sin x - (x - \frac{x^3}{3!} + \frac{x^5}{5!})}{x^5} = 0,$ and so $\lim_{x\to 0}\frac{\sin x - x + \frac{x^3}{3!}}{x^5} = \frac{1}{5!}.$ Finally, $\lim_{x \rightarrow 0}\frac{\cos x- 1 + \frac{x^2}{2}-\frac{x^4}{4!}}{x(\sin x - x+\frac{x^3}{3!})} = \lim_{x\to 0}\frac{\cos x - 1 + \frac{x^2}{2!} - \frac{x^4}{4!}}{x^6}\lim_{x\to 0}\frac{x^5}{\sin x - x + \frac{x^3}{3!}} = -\frac{1}{6}.$

Problem 5 Use Taylor’s theorem to calculate $\sqrt{1.01}$ with precision of at least $10^{-6}$.

Solution Let $f(x) = \sqrt{1+x}$. Note that $f'(x) = \frac{1}{2}(1+x)^{-1/2}, f''(x) = \frac{-1}{4}(1+x)^{-3/2}, f'''(x) = \frac{3}{8}(1+x)^{-5/2}.$ The 2nd order Taylor polynomial of $\sqrt{1+x}$ is $P(x) = f(0) + \frac{f'(0)}{1!}x - \frac{f''(0)}{2!}x^2 = 1 + \frac{1}{2}x - \frac{1}{8}x^2.$ By Taylor’s theorem, for some $0 < c < 0.01$ $|f(0.01) - P(0.01)| = \frac{f'''(c)}{3!}(0.01)^3 = \frac{(3/8)(1+c)^{-5/2}}{3!}10^{-6} < \frac{1}{16}10^{-6} < 10^{-7}.$ Therefore $f(0.01)$ is approximately $P(0.01)$ up to an error of $10^{-7}$.

Problem 6 Use Taylor’s theorem to prove the following inequality, for every $x \in \mathbb{R}$: $e^x \geq 1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!.$

Solution The 5th order Taylor polynomial of $e^x$ is $P(x) = 1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!,$ and $f^{(6)}(x) = e^x.$ By Taylor’s theorem, for some $c$ between $0$ and $x$, $e^x - P(x) = \frac{e^c}{6!}x^6,$ which is always $\ge 0$.

Concavity

Definition If the graph of $f$ lies above (below) all of its tangents on an interval $I$, then it’s called concave upward (downward) on $I$.

Concavity test How $f''$ helps determine the intervals of concavity? If $f''(x) > 0$ for all $x$ in $I$, then $f$ is concave upward on $I$. If $f''(x) < 0$ for all $x$ in $I$, then $f$ is concave downward on $I$.

Definition A point $P$ on curve $y = f(x)$ is called an inflection point if $f$ is continuous there and curve changes from concave upward to concave downward, or from concave downward to concave upward, at $P$.

Second derivative test Suppose $f''$ is continuous near $c$. If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $c$.

Example Let’s discuss the curve $y = f(x) = x^4 - 4x^3$ with respect to concavity, points of inflection and local minima and local maxima. Note that $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$ and $f''(x) = 12x^2 - 24x = 12x(x-2)$. The function is monotone decreasing on $(-\infty, 3)$ and increasing on $(3,\infty)$. Second derivative $f''(x) = 0$ gives $x = 0, 2$. Thus $(0, 0)$ and $(3, -27)$ are inflection points and $f$ is concave upward on $(-\infty, 0)$ and $(2, \infty)$, and it is concave downward on $(0,2)$. First derivative $f'(x) = 0$ gives $x = 0, 3$ are critical numbers. By the second derivative test, we know that $x = 3$ is a local minimum. However $x = 0$ is not local min/max because $f$ is monotone decreasing on $(-\infty, 3)$.