The limit lim _{x rightarrow infty} x^2 int _ | vedclass

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(A) We have,$\lim _{x \rightarrow \infty} x^2 \int _{0}^{x} e^{t^3-x^3} dt = \lim _{x \rightarrow \infty} x^2 e^{-x^3} \int _{0}^{x} e^{t^3} dt = \lim

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$\frac{1}{3}$

Vedclass · Answer

(A) We have,$\lim _{x \rightarrow \infty} x^2 \int _{0}^{x} e^{t^3-x^3} dt = \lim _{x \rightarrow \infty} x^2 e^{-x^3} \int _{0}^{x} e^{t^3} dt = \lim _{x \rightarrow \infty} \frac{x^2 \int _{0}^{x} e^{t^3} dt}{e^{x^3}}$.Since this is an $\frac{\infty}{\infty}$ form,we apply $L'Hospital's$ rule:$= \lim _{x \rightarrow \infty} \frac{\frac{d}{dx} (x^2 \int _{0}^{x} e^{t^3} dt)}{\frac{d}{dx} (e^{x^3})} = \lim _{x \rightarrow \infty} \frac{2x \int _{0}^{x} e^{t^3} dt + x^2 e^{x^3}}{3x^2 e^{x^3}} = \lim _{x \rightarrow \infty} \left( \frac{2 \int _{0}^{x} e^{t^3} dt}{3x e^{x^3}} + \frac{1}{3} \right)$.Applying $L'Hospital's$ rule again to the first term:$= \lim _{x \rightarrow \infty} \frac{2 e^{x^3}}{3(e^{x^3} + x \cdot 3x^2 e^{x^3})} + \frac{1}{3} = \lim _{x \rightarrow \infty} \frac{2 e^{x^3}}{3 e^{x^3}(1 + 3x^3)} + \frac{1}{3} = 0 + \frac{1}{3} = \frac{1}{3}$.

The limit $\lim _{x \rightarrow \infty} x^2 \int _{0}^{x} e^{t^3-x^3} dt$ equals

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