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Let f(x) = (xe)^(1/|x|+1/x); x != 0, f(...

Let `f(x) = (xe)^(1/|x|+1/x); x != 0, f(0) = 0`, test the continuity & differentiability at `x = 0`

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The correct Answer is:
(i)yes (ii) No
Given, ` f(x) = {{:(xe^(-(1/x+1/x))" , "x gt 0),(xe^(-(-1/x+1/x))" , "x lt 0),(" 0 , "x=0):}`
`{{:(xe^(2/x)" , "xgt0),(" x , "xlt0),(" 0 , "x = 0):}`
(i) To check continuity at x = 0 ,
LHL (at x =0)= ` underset( h to 0) lim - h = 0 `
RHL = ` underset( h to 0) lim h/(e^(2//h)) = 0`
Also, f(0) = 0
`:. ` f (x) is continuous at x = 0
(ii) To check differentiability at x = 0 ,
`L f'(0) = underset( h to 0) lim (f(0-h)-f(0))/(-h) `
` = underset( h to 0) lim lim ((0-h)-0)/(-h) = 1`
`R f'(0) = underset( h to 0) lim (f(0+h)-f(0))/h `
= ` underset( h to 0) lim (he^(-2//h)-0) /h =0`
`:. ` f(x) is not differentiable at x = 0.
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