What is the #lim_(x -> 4) (x-4)/(sqrtx - 2)#?

1 Answer
Jim H
Dec 1, 2015

It is #4#

Explanation:

Rationalize the denominator or factor the numerator.

Here is the factorization.

#x-4 = (sqrtx+2)(sqrtx-2)#, so we get:

#(x-4)/(sqrtx-2) = ((sqrtx+2)(sqrtx-2))/(sqrtx-2)#

So,
# (x-4)/(sqrtx-2) = sqrtx+2# for all #x != 4#

As #xrarr4#, we get #sqrtx+2 rarr 4#

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