$\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}$

solution:

Given $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}$

Now by splitting the limits in above equation we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{x}+\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}$

Taking constant term outside the limits we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{x}+\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x}-2(3) \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}+(5) \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}$

Taking constant term outside the limits we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{x}+\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x}-2(3) \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}+(5) \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}$

Now as $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

By substituting and applying the limits we get

$\Rightarrow$ $\lim _{x \rightarrow 0} \frac{\sin x}{x}-2(3) \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}+(5) \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}=1-6+5=0$

$\Rightarrow $ $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}=0$