Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 278 Question 278 of 401 Medium MCQ
If f ( x ) = { x e − ( 1 ∣ x ∣ + 1 x ) , x ≠ 0 0 , x = 0 f(x)=\begin{cases}xe^{-\left(\frac{1}{|x|}+\frac1x\right)}, & x\ne0\\0, & x=0\end{cases} f ( x ) = { x e − ( ∣ x ∣ 1 + x 1 ) , 0 , x = 0 x = 0 , then f ( x ) f(x) f ( x ) is
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RHL = lim h → 0 h e − 2 / h = 0 =\displaystyle\lim_{h\to0}he^{-2/h}=0 = h → 0 lim h e − 2/ h = 0 ; LHL = lim h → 0 ( − h ) e 0 = 0 =\displaystyle\lim_{h\to0}(-h)e^{0}=0 = h → 0 lim ( − h ) e 0 = 0 ; so f f f is continuous at 0 0 0 .
RHD = lim h → 0 h e − 2 / h h = 0 =\displaystyle\lim_{h\to0}\dfrac{he^{-2/h}}{h}=0 = h → 0 lim h h e − 2/ h = 0 ; LHD = lim h → 0 − h e 0 − h = 1 =\displaystyle\lim_{h\to0}\dfrac{-he^{0}}{-h}=1 = h → 0 lim − h − h e 0 = 1 . Since RHD ≠ \ne = LHD, f f f is not differentiable at x = 0 x=0 x = 0 . Answer the question to unlock