Answer
$0$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{1-\cos x}{\tan x+x (\sec^2 x)}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{\sin x}{2 \sec^2 x+2x\sec^2 x \tan x}=\dfrac{\sin 0}{2 \sec^2 0+0}$
or, $\dfrac{0}{2+0}=0$
