The value of limx→0 ∫0x cos t2dtx is
1
0
– 1
2
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.