Answer
$e^{1/2}$
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, $\ln f(x)=\ln (1+2x)^{1/2 \ln x} \implies\ln f(x)= \dfrac{\ln (1+2x)}{2 \ln x}$
Now, $\lim\limits_{x \to \infty} f(\infty)=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to \infty} \frac{2/1+2x}{2/ x}}=e^{\lim\limits_{x \to \infty} \frac{1}{1/ x+2}}$
Thus,
$e^{\lim\limits_{x \to \infty} \frac{1}{1/ x+2}}=e^{1/2}$