Question

Find the value of $\underset{x\to 0}{lim}\frac{e^{2x}+e^{-2x}-2}{e^x-x^{-x}}$

Answer 1

If we substitute x=0 in the given expression, we will get $\frac{0}{0}$.
We will look for same ways to factorize this. Numerator and denominator must be connected in some way to factorize.
In numerator we have $e^{2x}$ and $e^{-2x}$ and in denominator we have $e^{x}and{e}^{-x}$. So by squaring the denominator we may be able to get numerator.

$(e^x-e^{-x}{)}^2=e^{2x}+e^{-2x}-2\times e^x\times e^{-x}$

$=e^{2x}+e^{-2x}-2$

this is same as numerator.

$\Rightarrow \underset{x\to 0}{lim}\frac{e^{2x}+e^{-2x}-2}{e^x-e^{-x}}=\underset{x\to 0}{lim}\frac{(e^x-e^{-x}{)}^2}{e^x-e^{-x}}$

$=\underset{x\to 0}{lim}{e}^x-e^{-x}$

$=1-1$

$=0$