Investigate $f(x)=\frac{\tanh (x)-1}{e^{-2 x}}$ for $x \to \infty$

Question

The following are given

$$ \lim_{x \to \infty}{\log(x)} = \infty$$

$$ \lim_{x \to \infty}{\cosh(x)} = \infty$$

$$ \lim_{x \to \infty}{\sinh(x)} = \infty$$

$$ \tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \quad \text { for } x \in \mathbb{R} $$

$$ \lim_{x \to \infty}{x^2} = \infty$$

$$ \lim_{x \to \infty}{\exp(x)} = \infty$$

Both the denominator and the numerator goes to infinity. I do not know how to handle this. A hint or a bit more will be helpful.

Differentiation are not allowed.

Answer 1

\begin{align}\frac{\tanh (x)-1}{e^{-2 x}}&=e^{2x}\big(\tanh (x)-1\big)\\&= e^{2x}\left( \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}-1\right)\\&= e^{2x}\left(\frac{-2e^{-x}}{e^{x}+e^{-x}}\right)\\&= \frac{-2e^{x}}{e^{x}+e^{-x}}\\&=\frac{2}{e^{2x}+1}-2 \end{align}

Answer 2

$\tanh(x)-1=\frac{e^x-e^{-x}}{e^x+e^{-x}}-1=\frac{e^x-e^{-x}}{e^x+e^{-x}}-\frac{e^x+e^{-x}}{e^x+e^{-x}}=\frac{-2e^{-x}}{e^x+e^{-x}}$

$\implies \frac{\tanh(x)-1}{e^{-2x}}=\frac{-2e^{x}}{e^x+e^{-x}} \to -2$ as $x \to \infty$.