Question

In Problems 1– 6, find each limit

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

Short answer

The value is$\frac{2}{3}$

Step-by-step solution

Step 1. Distributing the limit

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

It can be written as

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

Step 2. Value of the limit

On simplifying the limit

$$\begin{gathered} \frac{{\lim }_{x\to \frac{\pi }{4}}\tan x}{{\lim }_{x\to \frac{\pi }{4}}\left(1+{\cos }^{2}x\right)}=\frac{{\lim }_{x\to \frac{\pi }{4}}\tan x}{{\lim }_{x\to \frac{\pi }{4}}1+{\lim }_{x\to \frac{\pi }{4}}{\cos }^{2}x} \\ =\frac{\tan \left(\frac{\pi }{4}\right)}{{\lim }_{x\to \frac{\pi }{4}}1+{\cos }^2\left(\frac{\pi }{4}\right)} \\ =\frac{\tan \left(\frac{\pi }{4}\right)}{1+{\cos }^2\left(\frac{\pi }{4}\right)} \\ =\frac{1}{1+{\left(\frac{\sqrt{2}}{2}\right)}^2} \\ =\frac{2}{3} \end{gathered}$$