lim _{x rightarrow frac{pi}{2}}left(frac{1}{l | vedclass

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(A) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\pi/2)^2}$.Since the limit is of the form $\frac{0}

Vedclass · Accepted Answer

$\frac{3 \pi^2}{8}$

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(A) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\pi/2)^2}$.Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule and use the Leibniz integral rule:$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx} \int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{2(x-\pi/2)}$$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{0 - \cos((x^3)^{1/3}) \cdot \frac{d}{dx}(x^3)}{2(x-\pi/2)}$$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos(x) \cdot 3x^2}{2(x-\pi/2)}$Let $h = x - \frac{\pi}{2}$,then $x = h + \frac{\pi}{2}$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.$L = \lim _{h \rightarrow 0} \frac{-\cos(h + \pi/2) \cdot 3(h + \pi/2)^2}{2h}$$L = \lim _{h \rightarrow 0} \frac{\sin(h) \cdot 3(h + \pi/2)^2}{2h}$Since $\lim _{h \rightarrow 0} \frac{\sin h}{h} = 1$,we get:$L = 1 \cdot \frac{3(\pi/2)^2}{2} = \frac{3 \cdot \pi^2/4}{2} = \frac{3 \pi^2}{8}$.

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

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