Chapter 4, Problem 34RE

To determine

Tofind:the limit of the function.

Answer to Problem 34RE

Thelimit of the function $\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ is $1$ .

Explanation of Solution

Given:

  $\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ .

Concept used:

If the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of the derivative.

Calculation:

  $\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ .

Let $l$ is the limit which is equal to the $\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ .

  $l=\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ .

By taking log to both side of the function.

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\text{logtanx}}{\text{secx}}$ .

By putting direct $x=\frac{\pi }{2}$ the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of the derivative.

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\text{logtanx}}{\text{secx}}$

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\frac{d}{dx}\left(\text{logtanx}\right)}{\frac{d}{dx}\left(\text{secx}\right)}$

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\frac{1}{\tan x}{\sec }^{2}x}{\sec \text{x}\tan x}$ .

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\sec x}{{\tan }^{2}x}$ .

By putting again $x=\frac{\pi }{2}$ the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\sec x}{{\tan }^{2}x}$ .

Finding derivative of numerator and denominator

  $\text{logl=}\underset{\text{x}\to {\frac{\text{π}}{\text{2}}}^{\text{-}}}{\text{lim}}\frac{\sec x\tan x}{2\tan x{\sec }^{2}x}=\frac{1}{2}\cos x=0.$

  $\text{logl=}0.$

  $l=1$ .

Hence the limit of the function $\underset{\text{x}\to {\frac{\pi }{2}}^{-}}{\text{lim}}{\left(\text{Tanx}\right)}^{\text{Cosx}}$ is $1$ .