If f(x)=⎧⎨⎩x1+exp(1/x),x≠00,x=0, then f(x) at x=0 is
A
Continuous
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B
Not continuous
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C
Differentiable
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D
Not differentiable
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Solution
The correct options are A Continuous C Not differentiable Given f(x)=⎧⎨⎩x1+exp(1/x),x≠00,x=0, Continuity at x=0 LHL=limh→00−h1+e1/(0−h) =limh→0−h1+e−1/(h)=−01=0 =limh→0h1+e1/(h)=01+∞=0 ∴LHL=RHL=f(0)=0 Hence it is continuous Differentiability at x=0 LHD=limh→0f(0−h)−f(0)0−h=limh→00−h1+e1/(0−h)−0−h=1 RHD=limh→0f(0+h)−f(0)0+h=limh→00+h1+e1/(0+h)−0h=0 ∴LHD≠RHD Hence it is not differentiable at x=0