Question

$\underset{x\to 0}{lim}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}$

Answer 1

$\underset{x\to 0}{lim}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}$

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

$=\underset{x\to 0}{lim}\frac{e^{\frac{2}{x}}-1}{e^{\frac{2}{x}}+1}\left(\frac{\infty }{\infty }form\right)$

Using L'Hospital Rule

$=\underset{x\to 0}{lim}\frac{e^{\frac{2}{x}}\times \frac{-2}{x^2}}{e^{\frac{2}{x}}\times \frac{-2}{x^2}}$

$=1$