Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.6

Solution:

### Part a

Note that $\cos x\sin x\ge -1$ and $\sin x\ge -1$, we have

$$

\lim_{x\to\infty}f(x)\ge \lim_{x\to\infty}(x-1)=\infty.

$$ Hence $\lim_{x\to\infty}f(x)=\infty$.

Similarly,

$$

\lim_{x\to\infty}g(x)\ge \lim_{x\to\infty}e^{-1}(x-1)=\infty.

$$ Hence $\lim_{x\to\infty}g(x)=\infty$.

### Part b

By product rule and the identity $\sin^2 x+\cos^2 x=1$, we have

\begin{align*}

f’(x)

=&\ 1+(-\sin x)\sin x+\cos x\cos x\\

=&\ 1-\sin^2 x+\cos^2 x\\

=&\ \cos^2 x+\cos^2 x\\

=&\ 2\cos^2 x.

\end{align*}

Similarly, by Chain rule and product rule

\begin{align*}

g’(x)=&\ (e^{\sin x}f(x))’\\

=&\ e^{\sin x}(\cos x) f(x)+e^{\sin x}f’(x)\\

=&\ e^{\sin x}(\cos x) f(x)+e^{\sin x} 2\cos^2 x\\

=&\ e^{\sin x}\cos x(f(x)+2\cos x).

\end{align*}

### Part c

Using Part b and by cancellation, we have

$$

\dfrac{f’(x)}{g’(x)}=\frac{2\cos^2 x}{e^{\sin x}\cos x(f(x)+2\cos x)}=\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}.

$$

### Part d

Clearly, we have

$$

|2e^{-\sin x}\cos x|\le 2e,\quad |2\cos x+f(x)|\ge f(x)-2.

$$

Since $\lim_{x\to \infty} f(x)=\infty$, we have

$$

\lim_{x\to\infty}\left|\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}\right|\le \lim_{x\to\infty}\frac{2e}{f(x)-2}=0.

$$ Hence

$$

\lim_{x\to\infty}\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}=0.

$$ However,

$$

\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty} e^{-\sin x}

$$ does not exist since $\lim_{x\to\infty}(-\sin x)$ does not exists.