Evaluate $ \lim_{x\to 0}|\frac{5^x - 5^{-x}}{5^x-1}| $ without using L'Hospital's rule.

Question

$$ \lim_{x\to 0}\left|\frac{5^x - 5^{-x}}{5^x-1}\right| $$

I know the limit is equal to 2. But I am not allowed to use L'Hospital. How can I evaluate the limit without L'Hospital?

Answer 1

Hint:

$$\lim_{x \to 0}\bigg\vert\frac{5^x-5^{-x}}{5^x-1}\bigg\vert = \lim_{x \to 0}\bigg\vert\frac{5^{2x}-1}{5^{2x}-5^x}\bigg\vert = \lim_{x \to 0}\bigg\vert\frac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}\bigg\vert$$

Answer 2

Set $y= 5^x$ and consider $y\longrightarrow 1$ $$\left|\frac{5^x - 5^{-x}}{5^x-1}\right| = \left|\frac{y-\frac{1}{y}}{y-1}\right| = \left|\frac{y^2-1}{y(y-1)}\right| = \left|\frac{y+1}{y}\right| \stackrel{y \to 1}{\longrightarrow} 2$$

Answer 3

Alternative approach: Recall that $\lim_{x\to0}\frac{a^x-1}{x}=\ln a$

$\lim_{x\to 0}\left|\frac{5^x - 5^{-x}}{5^x-1}\right|=\lim_{x\to 0}\left|\frac{5^{2x}-1 }{5^x(5^x-1)}\right|=\lim_{x\to 0} |\frac{5^{2x}-1 }{2x}|\cdot\frac{2}{5^x}\cdot|\frac{x}{5^x-1}|=\frac{2\ln5}{\ln5}=2 $

Answer 4

HINT

For positive $x$ we have: $\frac{5^x-5^{-x}}{5^x-1}= \frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$. As $x\rightarrow 0$ this approaches to $1+1=2$.