How do you find the limit $lim_(h->0)((4+h)^2-16)/h$ ?

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Answer 1 · 2014-09-20T12:02:49

$lim_{h to 0}{(4+h)^2-16}/h=8$

Let us look at some details.

$lim_{h to 0}{(4+h)^2-16}/h$

by multiplying out the numerator,

$=lim_{h to 0}{16+8h+h^2-16}/h$

by cancelling out $16$ 's,

$=lim_{h to 0}{8h+h^2}/h$

by factoring out $h$ from the numerator,

$=lim_{h to 0}{h(8+h)}/h$

by cancelling out $h$ 's,

$=lim_{h to 0}(8+h)=8+0=8$

How do you find the limit lim_(h->0)((4+h)^2-16)/hlim_(h->0)((4+h)^2-16)/h ?