# Evaluate the Given limit:

Question:

Evaluate the Given $\operatorname{limit} \lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x} a, b, a+b \neq 0$

Solution:

At $x=0$, the value of the given function takes the form $\frac{0}{0}$.

Now,

$\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}$

$=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin a x}{a x}\right) a x+b x}{a x+b x\left(\frac{\sin b x}{b x}\right)}$

$=\frac{\left(\lim _{a x \rightarrow 0} \frac{\sin a x}{a x}\right) \times \lim _{x \rightarrow 0}(a x)+\lim _{x \rightarrow 0} b x}{\lim _{x \rightarrow 0} a x+\lim _{x \rightarrow 0} b x\left(\lim _{b x \rightarrow 0} \frac{\sin b x}{b x}\right)} \quad[$ As $x \rightarrow 0 \Rightarrow a x \rightarrow 0$ and $b x \rightarrow 0]$

$=\frac{\lim _{x \rightarrow 0}(a x)+\lim _{x \rightarrow 0} b x}{\lim _{x \rightarrow 0} a x+\lim _{x \rightarrow 0} b x} \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$=\frac{\lim _{x \rightarrow 0}(a x+b x)}{\lim _{x \rightarrow 0}(a x+b x)}$

$=\lim _{1 \rightarrow 0}(1)$

$=1$