Evaluate the following limits
$\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}$, where $a, b, a+b \neq 0$
$=\lim _{x \rightarrow 0} \frac{\sin (a x)+b x}{a x+\sin (b x)}$
$=\lim _{x \rightarrow 0} \frac{\sin (a x)+b x}{a x+\sin (b x)} \times \frac{b x}{a x} \times \frac{a}{b}$
$=\lim _{x \rightarrow 0} \frac{\frac{\sin a x+b x}{a x}}{\frac{a x+\sin b x}{b x}} \times \frac{a}{b}$
$=\frac{a}{b} \times \frac{\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x}}{\lim _{x \rightarrow 0} \frac{a x+\sin b x}{b x}}$
$=\frac{a}{b} \times \frac{1+\frac{b}{a}}{1+\frac{a}{b}}$
$=\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{b}}{\mathrm{a}}$
$=1$
$\therefore \lim _{x \rightarrow 0} \frac{\sin (a x)+b x}{a x+\sin (b x)}=1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.