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  • <img alt="\mathrm{ \lim _{x \rightarrow 0}\left(\frac{1}{x^5} \int_0^x c^{-t^2} d t-\frac{1}{x^4}+\frac{1}{3 x^2}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Clim

\mathrm{ \lim _{x \rightarrow 0}\left(\frac{1}{x^5} \int_0^x c^{-t^2} d t-\frac{1}{x^4}+\frac{1}{3 x^2}\right)}

Option: 1

\frac{1}{5}


Option: 2

\frac{1}{10}


Option: 3

10


Option: 4

\frac{1}{6}


Answers (1)

best_answer

 Applying \mathrm{\lim _{x \rightarrow 0} \frac{3 \int_0^x e^{-t^2} d t-3 x+x^3}{3 x^5}\left(\frac{0}{0} form \right) }

\mathrm{=\lim _{x \rightarrow 0} \frac{3\left(e^{-x^2}\right)-3+3 x^2}{15 x^4} }

Applying L' Hospital Rule and Newton-Leibnitz.

Applying Newton -Leibnitz's formula so

\begin{aligned} \mathrm{\frac{d}{d x} \int_0^x e^{-t^2} d t} & \mathrm{=e^{-x^2} \frac{d}{d x}(x)-e^{-0} \cdot \frac{d}{d x}(0)} \\ & \mathrm{=e^{-x^2} \cdot 1-0=e^{-x^2} }\end{aligned}

Thus, \mathrm{\lim _{x \rightarrow 6} \frac{3 \int_0^1 e^{-t^2} d t-3 x+x^3}{3 x^5} }

\mathrm{=\lim _{x \rightarrow 0} \frac{3 e^{-x^2}-3+3 x^2}{15 x^4}\left(\frac{0}{0} \text { form }\right) }
Again differentiate w.r.to \mathrm{x} (apply L' Hospital Rule)

\begin{aligned} & \mathrm{=\lim _{x \rightarrow 0} \frac{3 e^{-x^2}(-2 x)+6 x}{60 x^3}=\lim _{x \rightarrow 0} \frac{-e^{-x^2}+1}{10 x^2} }\\ & \mathrm{=\lim _{x \rightarrow 0} \frac{1-e^{-x^2}}{10 x^2}=\frac{1}{10} }\end{aligned}
 

Posted by

Gautam harsolia

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