# Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{(\sin 2 x+3 x)}{(2 x+\sin 3 x)}$

Solution:

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

So, by using the above formula, we have

Divide numerator and denominator by $x$,

$\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\sin 3 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x+3 x}{x}}{\frac{2 x+\sin 3 x}{x}}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{x}+3}{2+\frac{\sin 3 x}{x}}=\lim _{x \rightarrow 0} \frac{\frac{2 \sin 2 x}{2 x}+3}{2+\frac{3 \sin 3 x}{3 x}}=\frac{2+3}{2+3}=1$

ALTER: by using the rule, Differentiate numerator and denominator

$\lim _{x \rightarrow 0} \frac{2 \cos 2 x+3}{2+3 \cos 3 x}=\frac{5}{5}=1$

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