Evaluate the following limits:

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is $\frac{0}{0}$

Formula used: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ and $\lim _{x \rightarrow 0} \frac{\tan x}{x}=1$

So, by using the above formula, we have

Divide numerator and denominator by $x$,

$\lim _{x \rightarrow 0} \frac{x \cos x+\sin x}{x^{2}+\tan x}=\lim _{x \rightarrow 0} \frac{\frac{x \cos x+\sin x}{x}}{\frac{x^{2}+\tan x}{x}}=\lim _{x \rightarrow 0} \frac{\cos x+\frac{\sin x}{x}}{x+\frac{\tan x}{x}}=\frac{1+1}{0+1}=2$

Therefore, $\lim _{x \rightarrow 0} \frac{x \cos x+\sin x}{x^{2}+\tan x}=2$