# 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{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}

By applying L'Hospital's Rule

\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}

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