Answer
$1$
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:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{2e^x(e^x-1)}{\sin x+x \cos x)}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{4e^{2x}-2e^x}{2 \cos x-x\sin x}=\dfrac{4e^{0}-2e^0}{2 \cos 0-0}$
or, $\dfrac{4-2}{2(1)-0}=1$
