Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 36

Determine the $\displaystyle \lim_{x \to \infty} \frac{e^x - e^{-x}-2x}{x- \sin x}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to \infty} \frac{e^x - e^{-x}-2x}{x- \sin x} = \frac{e^0 - e^{-0} - 2(0)}{0 - \sin (0)} = \frac{1-1-0}{0-0} = \frac{0}{0} \text{ Indeterminate}$

Thus by applying L'Hospital's Rule...
$\displaystyle \lim_{x \to \infty} \frac{e^x - e^{-x}-2x}{x- \sin x} = \lim_{x \to 0} \frac{e^x - e^{-x} (-1)-2}{1- \cos x} = \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x}$

If we evaluate the limit, we will still get indeterminate form, so by applying L'Hospital's Rule once more...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x} &= \lim_{x \to 0} \frac{e^x + e^{-x}(-1)}{-(-\sin x)}\\
\\
&= \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x}
\end{aligned}
\end{equation}
$


Again, by applying L'Hospital's Rule, since the limit is still indeterminate...
$\displaystyle \lim_{x \to \infty} \frac{e^x-e^{-x}}{\sin x} = \lim_{x \to 0} \frac{e^x - e^{-x}(-1)}{\cos x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \lim_{x \to 0} \frac{e^x + \frac{1}{e^x}}{\cos x} = \frac{e^0 + \frac{1}{e^0}}{\cos 0} = \frac{1+ \frac{1}{1}}{1} = \frac{2}{1} = 2$