The value of lim _{x rightarrow 0} frac{1}{x} | vedclass

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(A) Let $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.Using the property $

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$e^{\sin ^{2} y}$

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(A) Let $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.Using the property $\int_{b}^{a} f(t) dt = -\int_{a}^{b} f(t) dt$,we can rewrite the expression as:$L = \lim _{x}$ ${\rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t + \int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$$L = \lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$.Since this is a $\frac{0}{0}$ form,we apply $L$'$H$ôpital's Rule and the Leibniz Integral Rule:$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx} \int_{y}^{x+y} e^{\sin ^{2} t} d t}{\frac{d}{dx} (x)}$$L = \lim _{x}$ ${\rightarrow 0} \frac{e^{\sin ^{2}(x+y)} \cdot \frac{d}{dx}(x+y) - e^{\sin ^{2}(y)} \cdot \frac{d}{dx}(y)}{1}$$L = \lim _{x \rightarrow 0} e^{\sin ^{2}(x+y)} \cdot (1) - 0 = e^{\sin ^{2} y}$.

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

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