Question

Find the limits of $\frac{e^x-e^{-x}}{\log (1+x)}$, when $x=0$.

Answer 1

To find the limits of $\frac{e^x-e^{-x}}{\log (1+x)}$ when $x=0$

i.e.

$L=\underset{x\to 0}{lim}\frac{e^x-e^{-x}}{\log (1+x)}$

$=\underset{x\to 0}{lim}\frac{e^x-e^{-x}}{x}\times \frac{x}{\log (1+x)}$

$=\underset{x\to 0}{lim}\frac{(e^x-1)-(e^{-x}-1)}{x}\times \underset{x\to 0}{lim}\frac{x}{\log (1+x)}$

$=\{\underset{x\to 0}{lim}\frac{(e^x-1)}{x}+\underset{x\to 0}{lim}\frac{(e^{-x}-1)}{-x}\}\times \underset{x\to 0}{lim}\frac{x}{\log (1+x)}$

$=\{\underset{x\to 0}{lim}\frac{(e^x-1)}{x}+\underset{x\to 0}{lim}\frac{(e^{-x}-1)}{-x}\}\times \underset{x\to 0}{lim}\frac{1}{\frac{\log (1+x)}{x}}$

Using the formulae $\underset{x\to 0}{lim}\frac{(a^x-1)}{x}={\log }_{e}a$

$\therefore \underset{x\to 0}{lim}\frac{e^x-1}{x}=1$ and $\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$

Then we have,

$L=\{\underset{x\to 0}{lim}\frac{(e^x-1)}{x}+\underset{x\to 0}{lim}\frac{(e^{-x}-1)}{-x}\}\times \underset{x\to 0}{lim}\frac{1}{\frac{(1+x)}{x}}$

$=[1+1]\times 1$

$=2$