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Question #d66eb

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Feb 14, 2017

$lim_(h->0) ((x+h)^2-x^2)/h = 2x$

Explanation:

$lim_(h->0) ((x+h)^2-x^2)/h = lim_(h->0) (cancel(x^2) +2hx + h^2 -cancel(x^2))/h$

$lim_(h->0) ((x+h)^2-x^2)/h = lim_(h->0) (2cancelhx+h^cancel2)/cancelh$

$lim_(h->0) ((x+h)^2-x^2)/h = lim_(h->0) (2x+h)$

$lim_(h->0) ((x+h)^2-x^2)/h = 2x$

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