Limits with Trig

$\lim_{x\to 0} \frac{8x+\tan{x}}{\sin{x}}=8\lim_{x\to 0} \frac{x}{\sin{x}} +\lim_{x\to 0} \frac{\sin{x}}{\cos{x}\cdot \sin{x}}=8+1=9$

We can't simply plug in $x=0$ as it will be something undefined. So, the key technique here is L'Hopital's Rule .

Let

$f(x)=8x+\tan x$

$g(x)=\sin x$

Since

$\displaystyle \lim_{x \to 0} f(x)=0$

$\displaystyle \lim_{x \to 0} g(x)=0$

Then we have

$\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$

$=\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)}$

$=\displaystyle \lim_{x \to 0} \frac{8+\sec^2 x}{\cos x}$

Just plug in $x=0$ ,

$=9$