lim _{x} {rightarrow frac{pi}{2}} left( frac{ | vedclass

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Question

(A) Let $f(x) = \int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt$. Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpi

Vedclass · Accepted Answer

$\frac{9 \pi^2}{8}$

Vedclass · Answer

(A) Let $f(x) = \int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt$. Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's Rule.Using Leibniz Rule,$f'(x) = -(\sin(2x) + \cos(x)) \cdot 3x^2$.The limit becomes $\lim _{x \rightarrow \frac{\pi}{2}} \frac{-3x^2(\sin(2x) + \cos(x))}{2(x - \frac{\pi}{2})}$.As $x \rightarrow \frac{\pi}{2}$,let $h = x - \frac{\pi}{2}$,so $x = h + \frac{\pi}{2}$.$= \lim _{h \rightarrow 0} \frac{-3(h + \frac{\pi}{2})^2(\sin(2h + \pi) + \cos(h + \frac{\pi}{2}))}{2h}$.$= \lim _{h \rightarrow 0} \frac{-3(h + \frac{\pi}{2})^2(-\sin(2h) - \sin(h))}{2h}$.$= \lim _{h \rightarrow 0} \frac{3(h + \frac{\pi}{2})^2(\sin(2h) + \sin(h))}{2h}$.$= \frac{3(\frac{\pi}{2})^2}{2} \cdot \lim _{h}$ ${\rightarrow 0} (\frac{\sin(2h)}{h} + \frac{\sin(h)}{h}) = \frac{3\pi^2}{8} \cdot (2 + 1) = \frac{9\pi^2}{8}$.

$\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left( \frac{\int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt}{(x - \frac{\pi}{2})^2} \right)$ is equal to:

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