How do you evaluate the limit of $lim (2h^3-h^2+5h)/h$ as $h->0$ ?

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Answer 1 · 2017-05-24T05:08:32

$lim_(x->0)(2h^3-h^2+5h)/h=5$

$lim_(x->0)(2h^3-h^2+5h)/h$
$=lim_(x->0)(3*2h^2-2*h+5)/1$
$=5$

Both the numerator and denominator are $0$ when $h=0$ . We can apply the L'Hopital's rule to find the limit.
The limit is equal to the derivative of numerator over derivative of the denominator.

Answer 2 · 2017-05-24T07:03:39

The value of this limit is $5$ . See explanation.

$lim_{h->0}(2h^3-h^2+5h)/h=lim_{h->0}(h*(2h^2-h+5))/h=$

$=lim_{h->0} (2h^2-h+5) =5$

How do you evaluate the limit of lim (2h^3-h^2+5h)/hlim (2h^3-h^2+5h)/h as h->0h->0?