Chapter 8.2, Problem 44E

To determine

To calculate: The limit of the given expression $\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{\left(\cos x\right)}^{\cos x}$ .

Answer to Problem 44E

The limit of the function is 1.

Explanation of Solution

Given Information:

The expression is defined as,

  $\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{\left(\cos x\right)}^{\cos x}$

Formula Used:

L’hospital rule is defined as,

If the limit is giving undefined form $\frac{\infty }{\infty },\frac{0}{0}$ for the function $\frac{f\left(x\right)}{g\left(x\right)}$ .

  $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\frac{d}{dx}\left(f\left(x\right)\right)}{\frac{d}{dx}\left(g\left(x\right)\right)}$

Calculation:

Consider the given fraction,

  $\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{\left(\cos x\right)}^{\cos x}$

First find the limit of the expression.

  $\begin{array}{c}\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{\left(\cos x\right)}^{\cos x}=\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{e}^{\ln {\left(\cos x\right)}^{\cos x}}\\ =\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{e}^{\cos x\ln \left(\cos x\right)}\\ =e^{\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\cos x\ln \left(\cos x\right)}\end{array}$

Further simplify the limit and use L’hospital rule.

  $\begin{array}{c}\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\cos x\ln \left(\cos x\right)=\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\frac{\ln \left(\cos x\right)}{\sec x}\\ =\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\frac{\frac{-\sin x}{\cos x}}{\sec x\tan x}\left[\text{L'hospital rule}\right]\\ =\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\frac{-\tan x}{\sec x\tan x}\\ =-\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\cos x\end{array}$

Further simplify.

  $\begin{array}{c}\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }\cos x\ln \left(\cos x\right)=-\cos \left(\pi /2\right)\\ =0\end{array}$

Resubstitute the value of the obtained limit.

  $\begin{array}{c}\underset{x\to {\left(\pi /2\right)}^{-}}{\lim }{\left(\cos x\right)}^{\cos x}=e^0\\ =1\end{array}$

Hence, the limit of the function is 1.