The value of limx→0 ∫0x2 cos⁡t2dtxsin⁡x, is

A
3/2
B
1
C
-1
D
none of these

Solution:

We have,

$\begin{array}{l} lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos t^{2} d t}{x \sin x} \\ = lim_{x \to 0} \frac{2 x \cos x^{4}}{x \cos x + \sin x} [Using L'Hospital's Rule] \\ = lim_{x \to 0} \frac{2 \cos x^{4} - 8 x^{4} \sin x^{4}}{2 \cos x - x \sin x} = \frac{2 - 0}{2 - 0} = 1 \end{array}$

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