**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}<br />
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, <br />
\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).