Answer
$1$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} (e^{2x}-1)^x$
Let us consider that $y=(e^{2x}-1)^x \implies \ln y=x \ln (e^{2x}-1)=\dfrac{\ln (e^{2x}-1)}{1/x}$
But $\lim\limits_{x \to 0^{+}} \dfrac{\ln (e^{2x}-1)}{1/x}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0^{+}} \dfrac{\ln (e^{2x}-1)}{1/x} =\lim\limits_{x \to 0^{+}} \dfrac{-2x^2}{1-e^{-2x}} =\dfrac{\infty}{\infty}$
Again apply L-Hospital's rule to obtain:
$\lim\limits_{x \to 0^{+}} \dfrac{-2x^2}{1-e^{-2x}} =\dfrac{-4x}{2e^{-2x}}=0$
Thus, we have: $\lim\limits_{x \to 0^{+}} y=e^0=1$
