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: $\lim _{x \rightarrow 0} \frac{\tan x}{x}=1$ or we can used $L$ hospital Rule,

So, by using the above formula, we have

Divide numerator and denominator by $x$,

$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\tan x}=\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x-x}{x}}{\frac{3 x-\tan x}{x}}=\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x}{x}-1}{3-\frac{\tan x}{x}}=\lim _{x \rightarrow 0} \frac{\frac{2 \tan 2 x}{2 x}-1}{3-\frac{\tan x}{x}}=\frac{2-1}{3-1}=\frac{1}{2}$

Therefore, $\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\tan x}=\frac{1}{2}$