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The value of  limx00xcos t2dtx is 

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detailed solution

Correct option is A

Let f(x)=∫0x cos⁡ t2dt and g(x)=x. Then f(0)=g(0)=0.∴ limx→0 f(x)g(x)=limx→0 f′(x)g′(x)⇒ limx→0 ∫0x cos⁡ t2dtx=limx→0 cos⁡ x2⋅1−cos⁡ 0⋅01=limx→0 cos⁡ x21=cos⁡ 0=1.


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