Chapter 4.5, Problem 40E

To determine

The value of given limit.

Answer to Problem 40E

  $1$ .

Explanation of Solution

Given information:

A limit function is given as,

$\underset{x\to 0^{+}}{\lim }{\left(\tan 2x\right)}^x$ .

Concept used:

L'Hospital's Rule: If $f$ and $g$ are differentiable in an open interval containing $a$ and $f(a)=g(a)=0$ . Also, $g\text{'}(x)\ne 0$ (except possibly at $a$ ). Then-

  $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}$

This rule is also applicable when $f(a)=\pm \infty \text{ and }g(a)=\pm \infty$ .

Calculation:

Given limit function is, $\underset{x\to 0^{+}}{\lim }{\left(\tan 2x\right)}^x$ .

  $\begin{array}{l}\text{Suppose, }L=\underset{x\to 0^{+}}{\lim }f(x)\\ \Rightarrow L=\underset{x\to 0^{+}}{\lim }\left[{\left(\tan 2x\right)}^x\right]\\ \Rightarrow L=\underset{x\to 0^{+}}{\lim }{e}^{\ln \left[{\left(\tan 2x\right)}^x\right]}\left\{\text{Since, }u=e^{\ln u}\right\}\\ \Rightarrow L=\underset{x\to 0^{+}}{\lim }{e}^{2\cdot \ln \left(\tan 2x\right)}\\ \Rightarrow L=e^{\underset{x\to 0^{+}}{\lim }\left[2\cdot \ln \left(\tan 2x\right)\right]}\\ \text{Plug in the value }x=0,\\ \Rightarrow L=e^{\left[2\ln \left(\tan 0\right)\right]}\\ \Rightarrow L=e^0\\ \Rightarrow L=1\end{array}$

Therefore, $L=\underset{x\to 0^{+}}{\lim }\left[{\left(\tan 2x\right)}^x\right]=1$

Hence, value of the given limit function is $1$ .