Show that $$\frac{\tanh(x)-1}{e^{-2x}}$$ has a limit for $x \rightarrow \infty$ and the method to finding it.
I am guessing I have to use epsilon-delta-definitions for limits and then find a inequality that narrows the limit down. I am having such a hard time using the definitions to find the limit as it does not intuitively make sense to me and the ammount of notation throws me off as I can not use a generel method to solving such a problem.
$$L=\lim_{x \to \infty} \frac{\tanh x-1}{e^{-2x}}=\lim_{x\to \infty} \frac{-2e^{-x}}{e^{-2x}(e^{x}+e^{-x})}=\lim_{x\to \infty} \frac{-2}{(1+e^{-2x})}=-2.$$