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\(\lim_{x\rightarrow 1}(\sqrt[3]{x}-1)/(x-1)\)
Wolfram alpha did not explain this to me so well. How the heck do we end up with \(1/(1+\sqrt[3]{x}+{x}^{2/3})\)??
Please show me in great detail how to crack down something this complicated.
Thanks!

CurlyFry  Feb 5, 2018
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Find the following limit:

lim_(x->1) (x^(1/3) - 1)/(x - 1)

 

Simplify the expression inside the limit.

(x^(1/3) - 1)/(x - 1) = (x^(1/3) - 1)/(x - 1):

lim_(x->1) (x^(1/3) - 1)/(x - 1)

 

Rationalize the expression.

(x^(1/3) - 1)/(x - 1) = ((x^(1/3) - 1) (1 + x^(1/3) + x^(2/3)))/((x - 1) (1 + x^(1/3) + x^(2/3))) = 1/(1 + x^(1/3) + x^(2/3)):

lim_(x->1) 1/(1 + x^(1/3) + x^(2/3))

 

The limit of a continuous function at a point is just its value there.

lim_(x->1) 1/(1 + x^(1/3) + x^(2/3)) = 1/(1 + 1^(1/3) + 1^(2/3)) = 1/3:

 

=1/3

Guest Feb 5, 2018

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