Question

The value of $\underset{x\to 0}{lim}{\left(\frac{e^x+e^{-x}-2}{x^2}\right)}^{{1/x}^2}$

• A. $\frac{1}{12}$
• B. $\frac{1}{3}$
• C. $\frac{1}{11}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{12}$

${lim}_{x\to 0}{\left(\frac{e^x+e^{-x}-2}{x^2}\right)}^{\frac{1}{2}}$

$a^x=e^{\ln a^x}=e^{x\cdot \ln a}$

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

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

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

Apply L-Hospital's rule 5 times

$\therefore {lim}_{x\to 0}\left(\frac{2e^{2x}+3e^{2x}+1}{2(2e^{2x}{x}^2+14e^{2x}x+18e^{2x}+6-2x)}\right)$

$=\frac{2\times 0+3+1}{2(2\times 0+14\times 0+18+6-2\times 0)}$

$=\frac{4}{48}$

$=\frac{1}{12}$