Chapter 9.2, Problem 24E

To determine

To Find: The value of a limit using L'Hôpital's Rule, after identifying the indeterminate form.

Answer to Problem 24E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^x$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to c,\text{ then }\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $a\to 0^{+}$ ,

  $\sin a\to 0$ and $a\to 0$

Therefore, the limit is of the indeterminate form $0^0$ .

So, assume the limit is $L$ and take the natural logarithm of the equation $L=\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^x$ so that

  $\begin{array}{l}L=\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^x\\ \Rightarrow \ln L=\ln \left[\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^x\right]\\ \Rightarrow \ln L=\underset{x\to 0}{\lim }\left[\ln {\left(\sin x\right)}^x\right]\\ \Rightarrow \ln L=\underset{x\to 0}{\lim }\left[x\ln \left(\sin x\right)\right]\\ \Rightarrow \ln L=\underset{x\to 0}{\lim }\left[\frac{\ln \sin x}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right]\end{array}$

Now, the righthand side of the above equation is in $\frac{\infty }{\infty }$ form.

So, apply L'Hôpital's Rule on the righthand side of the equation to write

  $\begin{array}{l}\ln L=\underset{x\to 0}{\lim }\frac{\left(\raisebox{1ex}{$\cos x$}\!\left/ \!\raisebox{-1ex}{$\sin x$}\right.\right)}{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}\\ =\underset{x\to 0}{\lim }\frac{-x^2}{\tan x}\left[\text{Again apply L'Hôpital's Rule because the limit is still in }\frac{0}{0}\text{ form}\right]\\ \underset{x\to 0}{=\lim }\frac{-2x}{{\sec }^{2}x}\\ =\frac{-2\times 0}{{\sec }^{2}0}\\ =0\\ \Rightarrow L=e^0=1\end{array}$

Therefore, the value of the limit is $1$ .