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Question

# Let f(x)=(tanxx)1x2x≠0,f(0)=0 .Then

A
f is continuous at x=0
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B
limx0f(x)=e13
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C
f is discontinuous at x=0
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D
limx0f(x)=0
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Solution

## The correct option is B limx→0f(x)=e13f(x)=(tanxx)1x2limx→0f(x)=limx→0(tanxx)1x2=limx→0explog(tanxx)1x2=explimx→0log(tanxx)x2=explimx→0x−tanx+xtan2x2x2tanx (Applying L-Hospital's rule)=explimx→0tanx(1+tan2x)x+2tanx+xtan2x (Again applying L-Hospital's rule)=exp((limx→0(1+tan2x))(limx→0tanxx+2tanx+xtan2x))=exp(limx→0tanxx+2tanx+xtan2x)=exp(limx→0sec2x1+2sec2x+tan2x+2xtanxsec2x) (Applying L-Hospital's rule)=e13

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