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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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Verified Answer
The correct answer is:
$e^{\sin ^{2} y}$
We have
$\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$
$=\lim _{x \rightarrow 0} \frac{\int_y^{x+y}e^{sin^{2t}}dt}{x}$
$=\lim _{x \rightarrow 0} \frac{e^{\sin ^{2}(x+y)}(1+0)-0}{1}=e^{\sin ^{2} y}$
$\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$
$=\lim _{x \rightarrow 0} \frac{\int_y^{x+y}e^{sin^{2t}}dt}{x}$
$=\lim _{x \rightarrow 0} \frac{e^{\sin ^{2}(x+y)}(1+0)-0}{1}=e^{\sin ^{2} y}$
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