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