Chapter 14, Problem 6CT

To determine

To calculate: The $\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan x}{1+{\cos }^{2}x}$.

Answer to Problem 6CT

Solution:

The $\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan x}{1+{\cos }^{2}x}$ is $\frac{2}{3}$.

Explanation of Solution

Given information:

$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan x}{1+{\cos }^{2}x}$.

Formula used:

Limit of a quotient: $\underset{x\to c}{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to c}{\lim }f\left(x\right)}{\underset{x\to c}{\lim }g\left(x\right)}$ Provided that $\underset{x\to c}{\lim }g\left(x\right)\ne 0$

Calculation:

The limit of denominator $\underset{x\to \frac{\pi }{4}}{\lim }\left(1+{\cos }^{2}x\right)=1+{\cos }^2\left(\frac{\pi }{4}\right)=1+{\left(\frac{\sqrt{2}}{2}\right)}^2=1+\frac{1}{2}=\frac{3}{2}$

As the limit of denominator is not zero so, use the limit of quotient rule

Thus, $\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan x}{1+{\cos }^{2}x}=\frac{\underset{x\to \frac{\pi }{4}}{\lim }\left(\tan x\right)}{\underset{x\to \frac{\pi }{4}}{\lim }\left(1+{\cos }^{2}x\right)}$

$=\frac{\tan \left(\frac{\pi }{4}\right)}{1+{\cos }^2\left(\frac{\pi }{4}\right)}$

$=\frac{1}{1+{\left(\frac{\sqrt{2}}{2}\right)}^2}$

$=\frac{1}{1+\frac{1}{2}}$

$=\frac{1}{\frac{3}{2}}$

$=\frac{2}{3}$

Thus the $\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan x}{1+{\cos }^{2}x}$ is $\frac{2}{3}$.