lim _{x rightarrow 0} frac{1}{x}left[int_{y}^ | vedclass

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Question

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

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

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(A) ધારો કે $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.ગુણધર્મ $\int_{b}^{a} f(t) dt = -\int_{a}^{b} f(t) dt$ નો ઉપયોગ કરતા:$L = \lim _{x}$ ${\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]$$L = \lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$.આ $\frac{0}{0}$ સ્વરૂપ હોવાથી, $L$'$H$ôpital ના નિયમ અને Leibniz સંકલન નિયમનો ઉપયોગ કરતા:$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx} \int_{y}^{x+y} e^{\sin ^{2} t} d t}{\frac{d}{dx} (x)}$$L = \lim _{x \rightarrow 0} \frac{e^{\sin ^{2}(x+y)} \cdot (1) - 0}{1} = e^{\sin ^{2} y}$.

$\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ ની કિંમત શોધો.

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