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Question Number 216491 by Jubr last updated on 09/Feb/25

Answered by MrGaster last updated on 09/Feb/25

$$\left(1\right):$$$$=\underset{x\to 0}{\lim }\frac{x{\int }_0^x(1-t^2+\frac{t^4}{2!}-\frac{t^6}{3!}+\dots )dt}{\sqrt{1-e^{-x^2}}}$$$$=\underset{x\to 0}{\lim }\frac{x(x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{3!}+\dots )dt}{\sqrt{1-e^{-x^2}}}$$$$=\underset{x\to 0}{\lim }\frac{x^2-\frac{x^4}{3}+\frac{x^6}{10}-\frac{x^8}{42}+\dots }{\sqrt{1-e^{-x^2}}}$$$$=\underset{x\to 0}{\lim }\frac{x^2}{\sqrt{1-(1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\dots )}}$$$$=\underset{x\to 0}{\lim }\frac{x^2}{\sqrt{1-\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^6}{24}-\dots }}$$$$=\underset{x\to 0}{\lim }\frac{x\sqrt{2}}{\sqrt{1-\frac{x^2}{4}+\frac{x^4}{24}-\dots }}$$$$=\underset{x\to 0}{\lim }x\sqrt{2}$$$$=\overline{)\begin{array}{c}0\end{array}}$$$$\left(2\right):$$$${\int }_0^x{e}^{-t^2}dt\approx x-\frac{x^3}{3}+\frac{x^5}{10}-\dots$$$$x{\int }_0^x{e}^{-t^2}dt\approx x^2-\frac{x^4}{3}+\frac{x^6}{10}-\dots$$$$\sqrt{1-e^{-x^2}}\approx \sqrt{x^2-\frac{x^4}{2}+\dots }\approx x(1-\frac{x^2}{4}+\dots )$$$$\frac{x{\int }_0^x{e}^{-t^2}dt}{\sqrt{1-e^{-x^2}}}\approx \frac{x^2-\frac{x^4}{3}}{x(1-\frac{x^2}{4})}\approx \frac{x(1-\frac{x^2}{3})}{1-\frac{x^2}{4}}$$$$\approx x(1-\frac{x^2}{12})\to \overline{)\begin{array}{c}0\end{array}}$$

Commented by Jubr last updated on 09/Feb/25

$$Thankssir.$$

Answered by mathmax last updated on 10/Feb/25

$$u\left(x\right)=x{\int }_0^x{e}^{-t^2}dtetv\left(x\right)=\sqrt{1-e^{-x^2}}$$$${lim}_{x\to 0}u\left(x\right)={lim}_{x\to 0}v\left(x\right)=0onappliquelhospital$$$$l={lim}_{x\to 0}\frac{u^{{}^{\prime }}\left(x\right)}{v^{{}^{\prime }}\left(x\right)}={lim}_{x\to 0}\frac{{\int }_0^x{e}^{-t^2}dt+{xe}^{-x^2}}{\frac{2{xe}^{-x^2}}{2\sqrt{1-e^{-x^2}}}}$$$$={lim}_{x\to 0}\frac{\sqrt{1-e^{-x^2}}({\int }_0^x{e}^{-t^2}dt+{xe}^{-x^2})}{{xe}^{-x^2}}$$$$wehave{e}^{-x^2}\sim 1-x^2\Rightarrow 1-e^{-x^2}\sim x^2$$$$and\sqrt{1-e^{-x^2}}\sim x\Rightarrow$$$$l={lim}_{x\to 0}\frac{x({\int }_0^x{e}^{-t^2}dt+{xe}^{-x^2})}{{xe}^{-x^2}}$$$$={lim}_{x\to 0}\frac{{\int }_0^x{e}^{-t^2}dt+{xe}^{-x^2}}{e^{-x^2}}=\frac{o}{1}=0$$

Commented by Jubr last updated on 10/Feb/25

$$Thankssir.$$

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