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  • Evaluate  <img alt="\mathrm{\lim _{x \rightarrow \infty}\left(\int_0^{\pi / 6}(\sin t)^x d t\right)^{1 / x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20

Evaluate  \mathrm{\lim _{x \rightarrow \infty}\left(\int_0^{\pi / 6}(\sin t)^x d t\right)^{1 / x}}

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{4}


Option: 3

0


Option: 4

1


Answers (1)

best_answer

Note that for every \mathrm{x>\frac{6}{\pi}}, when \mathrm{\frac{\pi}{6}-\frac{1}{x} \leq t \leq \frac{\pi}{6}, 0<\sin \left(\frac{\pi}{6}-\frac{1}{x}\right) \leq \sin t \leq \frac{1}{2}}. It follows that

                               \mathrm{ \sin \left(\frac{\pi}{6}-\frac{1}{x}\right)\left(\frac{1}{x}\right)^{\frac{1}{x}} \leq\left(\int_0^{\frac{\pi}{6}}(\sin t)^x d t\right)^{\frac{1}{x}} \leq \frac{1}{2}\left(\frac{\pi}{6}\right)^{\frac{1}{x}} . }

Letting \mathrm{x \rightarrow \infty}, it follows that

                                          \mathrm{ \lim _{x \rightarrow \infty}\left(\int_0^{\frac{x}{6}}(\sin t)^x d t\right)^{\frac{1}{x}}=\frac{1}{2} }

Posted by

rishi.raj

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