Limit Exercise using L'Hopital Rule and FTC

Exercise:

Compute , $\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{3}}}\int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt}$

Solution: as $\frac{{{t}^{2}}}{{{t}^{4}}+1}$ is continuous at $0$ then $\underset{x\to 0}{\mathop{\lim }}\,\int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt}=0$

Hence $\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{3}}}\int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt}=\frac{0}{0}\,ind\,form$ Apply L’Hopital Rule we get

$\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{3}}}\int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3{{x}^{2}}}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{d}{dx}\left( \int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt} \right)$ by F.T.C

We get $\frac{d}{dx}\left( \int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt} \right)=\frac{{{x}^{2}}}{{{x}^{4}}+1}$

Hence $\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{3}}}\int_{0}^{x}{\frac{{{t}^{2}}}{{{t}^{4}}+1}dt}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3{{x}^{2}}}\times \frac{{{x}^{2}}}{{{x}^{4}}+1}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3}\times \frac{1}{{{x}^{4}}+1}=\frac{1}{3}$