How do you find the limit of $\frac{\sqrt{x} - 2}{x - 4}$ $(sqrtx-2)/(x-4)$ as $x \to 4^{-}$ $x->4^-$ ?

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Answer 1 · 2016-12-13T04:16:31

$lim_(x->4^-)(sqrtx-2)/(x-4)=1/4$

Since the function,
as $x->4^-$ , is equal to the indeterminate form: $0/0$

Therefore:

$lim_(x->4^-)(sqrtx-2)/(x-4)=>0/0$

$=^(L'H) lim_(x->4^-)(1/2x^(-1/2))=1/(2sqrt4)=1/4$

How do you find the limit of (sqrtx-2)/(x-4)√x−2x−4(sqrtx-2)/(x-4) as x->4^-x→4−x->4^-?