Let f: left[-frac{pi}{2}, frac{pi}{2}right] r | vedclass

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(B) We are given the limit $\alpha = \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{e^{x^2}-1}$.We can rewrite the expression as $\alpha = \lim _{x

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(B) We are given the limit $\alpha = \lim _{x ightarrow 0} \frac{x \int_0^x f(t) dt}{e^{x^2}-1}$.We can rewrite the expression as $\alpha = \lim _{x ightarrow 0} \frac{\int_0^x f(t) dt}{x} \cdot \frac{x^2}{e^{x^2}-1}$.Since $\lim _{x ightarrow 0} \frac{e^{x^2}-1}{x^2} = 1$,we have $\lim _{x ightarrow 0} \frac{x^2}{e^{x^2}-1} = 1$.Now,applying $L$'$H$ôpital's rule to the first part $\lim _{x ightarrow 0} \frac{\int_0^x f(t) dt}{x}$ (which is of the form $\frac{0}{0}$):$\lim _{x ightarrow 0} \frac{\frac{d}{dx} \int_0^x f(t) dt}{\frac{d}{dx} x} = \lim _{x ightarrow 0} \frac{f(x)}{1} = f(0)$.Given $f(0) = \frac{1}{2}$,we get $\alpha = \frac{1}{2} 	imes 1 = \frac{1}{2}$.Finally,we calculate $8 \alpha^2 = 8 	imes \left(\frac{1}{2}ight)^2 = 8 	imes \frac{1}{4} = 2$.

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=\frac{1}{2}$. If $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{e^{x^2}-1}=\alpha$,then $8 \alpha^2$ is equal to:

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