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).