Hello, please consider the following.
[tex]\displaystyle \forall x \in \mathbb{R}\\\\\dfrac{e^{2x}-1}{e^x-1}\\\\=\dfrac{(e^x)^2-1^2}{e^x-1}\\\\=\dfrac{(e^x-1)(e^x+1)}{e^x-1}\\\\=e^x+1\\\\\text{So, we can find the limit.}\\\\\lim_{x\rightarrow 0} \ {\dfrac{e^{2x}-1}{e^x-1}}\\\\=\lim_{x\rightarrow 0} \ {e^x+1}\\\\=e^0+1\\\\\large \boxed{\sf \bf \ =2 \ }[/tex]
Thank you
