Question

$\underset{x\to 0}{\mathrm{Lim}}\frac{\underset{0}{\overset{x}{\int }}{\left(t^2+e^{t^2}\right)}^{\frac{1}{1-\cos t}}dt}{\left(e^x-1\right)}$ is equal to

• A. $e^4$
• B. $e^2$
• C. $e^3$
• D. $e$

Answer 1

Answer: (A)

$L=\underset{x\to 0}{\mathrm{Lim}}\frac{\underset{0}{\overset{x}{\int }}{\left(t^2+e^{t^2}\right)}^{\frac{1}{1-\cos t}}dt}{x}\left(\frac{0}{0}\right)$