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  • Evaluate <img alt="\mathrm{\lim _{x \rightarrow \infty} x e^{x^2}\left(\int_0^x e^{-t^2} d t-\int_0^{\infty} e^{-t^2} d t\right)}" src="/latex-image/?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow

Evaluate \mathrm{\lim _{x \rightarrow \infty} x e^{x^2}\left(\int_0^x e^{-t^2} d t-\int_0^{\infty} e^{-t^2} d t\right)}

Option: 1

1

 


Option: 2

2


Option: 3

-\frac{1}{2}


Option: 4

5


Answers (1)

best_answer

Using integration by parts, 

           \mathrm{\int_x^{\infty} e^{-t^2} \mathrm{~d} t=\int_x^{\infty} \frac{2 t e^{-t^2}}{2 t} \mathrm{~d} t=\frac{e^{-x^2}}{2 x}-\int_x^{\infty} \frac{e^{-t^2}}{2 t^2} \mathrm{~d} t,}

and 

                        \mathrm{\int_x^{\infty} \frac{e^{-t^2}}{2 t^2} \mathrm{~d} t=\frac{e^{-x^2}}{4 x^3}-\int_x^{\infty} \frac{3 e^{-t^2}}{4 t^4} \mathrm{~d} t .}

Since the integrands are always non-negative, this gives

                       \mathrm{\frac{e^{-x^2}}{2 x}-\frac{e^{-x^2}}{4 x^3} \leq \int_x^{\infty} e^{-t^2} \mathrm{~d} t \leq \frac{e^{-x^2}}{2 x}}

Thus, 

                     \mathrm{x e^{x^2}\left(\int_0^x e^{-t^2} \mathrm{~d} t-\int_0^{\infty} e^{-t^2} \mathrm{~d} t\right)=-x e^{x^2} \int_x^{\infty} e^{-t^2} \mathrm{~d} t=-\frac{1}{2}+O\left(x^{-2}\right)}

and the limit is -\frac{1}{2}

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Gunjita

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