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 a limit, so we need to use L'Hospital's rule.
$ \lim\limits_{x \to 0^{+}} \dfrac{2x+2/x^2+2x}{1/x}=\lim\limits_{x \to 0^{+}} \dfrac{2x^2+2x}{x^2+2x}=\dfrac{0}{0}$
Again apply L'Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{4x+2}{2x+2}=\dfrac{2}{2}=1$
