Limit $\lim_{x\to+\infty} {e^{2x}−1\over e^x−1}$

Hint: Let $u=e^x$. We then have

$$\lim_{x\to+\infty} {e^{2x}−1\over e^x−1}=\lim_{u\to+\infty} {u^2−1\over u−1}=\lim_{u\to\infty}\frac{(u+1)(u-1)}{u-1}$$

Perhaps this is even simpler $$ \lim_{x\to+\infty} {e^{2x}−1\over e^x−1}=\lim_{x\to+\infty} {e^{2x}−1\over e^x−1}\left({e^{-x}\over e^{-x}}\right)=\lim_{x\to+\infty} {e^{x}−e^{-x}\over 1−e^{-x}}={\infty - 0 \over 1-0}=\infty $$

observe that for $x>0$ we have $$\frac{e^{2x}-1}{e^x-1}>e^x$$ so the answer is $\infty$