Finding limit of hyperpolic expression.

Question

I am having trouble of how to solve this kind of problem. I have to show the limit of the function:

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

$\lim_{x\to\infty} f(x)$

I am to do this without using differential calculus. How do i go about doing this?

EDIT:

$\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

Answer 1

use that $$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and we get $$\frac{1-\tanh(x)}{e^{-2x}}=\frac{2e^{-x}}{e^{-2x}(e^x+e^{-x})}=\frac{2e^{x}}{e^x+e^{-x}}=\frac{2}{1+e^{-2x}}$$