Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 54

Determine the $\displaystyle \lim_{x \to 0^+} (\tan 2x)^x$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

If we let $y = (\tan 2x)^x$, then
$\ln y = x \ln [\tan (2x)]$

So,

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0^+} \ln y &= \lim_{x \to 0^+} x \ln [\tan (2x)]\\
\\
&= \lim_{x \to 0^+} \frac{\ln[\tan (2x)]}{\frac{1}{x}}
\end{aligned}
\end{equation}
$


By applying L'Hospital's Rule

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0^+} \frac{\ln[\tan (2x)]}{\frac{1}{x}} &= \lim_{x \to 0^+} \frac{\frac{2 \sec^2 2x}{\tan 2x}}{\frac{-1}{x}} \\
\\
&= \lim_{x \to 0^+} \frac{\frac{2 \left( \frac{1}{\cos^2 2x}\right)}{\frac{\sin 2x}{\cos 2x}} }{-\frac{1}{x^2}}\\
\\
&= \lim_{x \to 0^+} \frac{\frac{2}{\sin 2x \cos 2x}}{\frac{-1}{x^2}}\\
\\
&= \lim_{x \to 0^+} \frac{2x^2}{\sin 2x \cos 2x}\\
\\
\text{Recall that } \lim_{x \to 0^+} \frac{\sin x}{x} &= 1, \text{ then}\\
\\
\lim_{x \to 0^+} \frac{-2x^2}{\sin 2x \cos 2x} &= -\left[ \lim_{x \to 0^+} \left( \frac{2x}{\sin 2x} \right) \cdot \lim_{x \to 0^+} \left( \frac{x}{\cos 2x} \right)\right]\\
\\
&= - \left[ 1 \cdot \frac{0}{\cos 2(0)} \right] = -[1 \cdot 0] = 0
\end{aligned}
\end{equation}
$