Justify $e^{2x}\log{x}\sim (x-1)$

Question

How do I justify that for $x\rightarrow1$

$$e^{2x}\log{x}\sim (x-1)$$

I know that if $x\rightarrow 0$, then $\log{x}\sim(x-1)$ because $\log(1+x)\sim x$ for $x\rightarrow0$. This limit does not approach $0$, though, but 1. Any hints?

This is what my textbook does:

$$\frac{e^{2x}\log{x}}{(x-1)^{1/3}}\sim\frac{(x-1)}{(x-1)^{1/3}}$$

Answer 1

As $x\rightarrow 1$, we have $\log x\sim x-1$, and $e^{2x}\sim e^2$, so $e^{2x}\log x\sim e^2(x-1)$.

Answer 2

By L'Hospital we have

$\lim_{x \to 1}\frac{e^{2x}\log{x}}{x-1}= \lim_{x \to 1}(2e^{2x} \log x+e^{2x}\frac{1}{x})=e^2$.