Answer
$1$
Work Step by Step
L'-Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{2e^x(e^x-1)}{\sin x+x \cos x)}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{4e^{2x}-2e^x}{2 \cos x-x\sin x}=\dfrac{4-2}{2(1)-0}=1$
