Find $\lim_{x\to \infty}\frac{1-\tanh(x)}{e^{-2x}}$ without L'Hopital

Question

I have a problem with solving limits of hyperbolic functions. Since I am not allowed to use l'Hopital I know that I have to change the fraction so I don't get a $0/0$ or $\infty/\infty$. But my problem is that I have tried lots of different things that doesn't seem to work, so I could really use a hint or two.

$$\lim_{x\to \infty}\frac{1-\tanh(x)}{e^{-2x}}$$

I've tried replacing $\tanh(x)$ with $\frac{e^x-e^{-x}}{e^x+e^{-x}}$ but since this fraction also gives me an $\frac{\infty}{\infty}$ it doesn't seem to help me.

Answer 1

You correctly substituted $\tanh(x)$ with $\frac{e^{x}-e^{-x}}{e^x+e^{-x}}$ and so now you only need to proceed and rewrite the numerator of the fraction as $$1-\tanh(x)=1-\frac{e^{x}-e^{-x}}{e^x+e^{-x}}=\frac{e^x+e^{-x}-e^x+e^{-x}}{e^x+e^{-x}}=\frac{2e^{-x}}{e^{x}+e^{-x}}=\frac{\not e^{x}}{\not e^x}\frac{2e^{-2x}}{1+e^{-2x}}$$ Returning to the initial limit this gives you $$\lim_{x\to+\infty}\frac{1-\tan(h)}{e^{-2x}}=\lim_{x\to+\infty}\dfrac{\dfrac{2\not e^{-2x}}{1+e^{-2x}}}{\not e^{-2x}}=\lim_{x\to+\infty}\frac{2}{1+e^{-2x}}=\frac{2}{1+0}=2$$

Answer 2

Hint: $$ \lim_{x\to\infty}\frac{1-\tanh(x)}{e^{-2x}} = \lim_{x\to\infty}2-\frac{2}{e^{2x}+1} $$ The calculations maybe will be easy doing $e^x = t$: $$ \frac{1-\tanh(x)}{e^{-2x}} = \frac{1-(t-1/t)/(t+1/t)}{1/t^2} = \cdots $$

Answer 3

Your initial thoughts were good;

\begin{align*} \lim_{x\to \infty} \frac{1-\tanh x}{e^{-2x}} &= \lim_{x\to\infty}\frac{1-\displaystyle\frac{e^x-e^{-x}}{e^x+e^{-x}}}{e^{-2x}}\\ &=\lim_{x\to\infty}\frac{\displaystyle\frac{e^{x}+e^{-x}-e^{x}+e^{-x}}{e^x+e^{-x}}}{e^{-2x}}\\ &=\lim_{x\to\infty}\frac{\displaystyle\frac{2e^{-x}}{e^{x}+e^{-x}}}{e^{-2x}}\\ &=\lim_{x\to\infty}\frac{\displaystyle\frac{2e^{-x}}{e^{x}+e^{-x}}}{e^{-2x}}\cdot\frac{e^{2x}}{e^{2x}}\\ &=\lim_{x\to\infty}\frac{\displaystyle\frac{2e^{-x}e^{2x}}{e^{x}+e^{-x}}}{1}\\ &=\lim_{x\to\infty}\frac{2e^x}{e^x+e^{-x}}\cdot\frac{e^{-x}}{e^{-x}}\\ &=\lim_{x\to\infty}\frac{2}{1+e^{-2x}}\\ &=\lim_{x\to\infty}\frac{2}{1+\frac{1}{e^{2x}}}\\ &=\frac{2}{1+0}\\ &= 2 \end{align*}