`g(x) = sqrt(|x|), c = 0` Use the alternate form of the derivative to find the derivative at x = c (if it exists)

Textbook Question

Chapter 2, 2.1 - Problem 69 - Calculus of a Single Variable (10th Edition, Ron Larson).
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leonard-chen's profile pic

leonard-chen | (Level 2) Adjunct Educator

Posted on

`lim_(x->0) (g(x)-g(c))/(x-c)`

`lim_(x->0) (sqrt(|x|))/x`

` ` In order to evaluate the above function, both the left hand limit and right hand limit need to be evaluated.

`L.H.L. = lim_(x->0^-) (sqrt(|x|))/(x) = lim_(x->0^-) (sqrt(-x))/(x) = DNE`

` `

`R.H.L. = lim_(x->0^+) (sqrt(|x|))/(x) = lim_(x->0^+) (sqrt(x))/(x) = DNE`

``Because the L.H.L. and R.H.L. both do not exist, the limit at x=0 does not exist, and therefore the derivative at x=0 does not exist.

` `

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`g(x)=sqrt(abs x) ` 

`g'(x)=lim_(h->0) (f(x+h)-f(x))/h`

`g'(x)=lim_(h->0) ((x+h)^(1/2)-(x)^(1/2))/h`

at x=0

`g'(x)=lim_(h->0) h^(1/2)/h`

`g'(x)=lim_(h->0) 1/h^(1/2)`

So, the limit is undefined , g'(x) does not exists.

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