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 Evaluate limx0xex2dx20xe2x2dx

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detailed solution

Correct option is A

Since ex2>0,e2x2>0 in [0,x] , where x>0 , ∫0x ex2dx and ∫0x e2x2dx→∞ as x→∞L=limx→∞ ∫0x ex22∫0x e2x2 is of the form ∞∞ .  Therefore, using L' Hopital's rule L=limx→∞ 2∫0x ex2dxddx∫0x ex2e2x2 =limx→∞2 ex2∫0x ex2dx ex22 =2limx→∞ ∫0x ex2dxex2∞∞ form    (Using L'Hopital Rule)  =2limx→∞ ex22xex2   =limx→∞ 1x=0


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