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:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
Then, $ \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}$
Now, again apply L-Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{4x+2}{2x+2}=\dfrac{4(0)+2}{2(0)+2}=1$
