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

Question

4668 views around the world

You can reuse this answer
Creative Commons License

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#