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

2 Answers
May 24, 2017

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

Explanation:

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

  1. Both the numerator and denominator are #0# when #h=0#. We can apply the L'Hopital's rule to find the limit.
  2. The limit is equal to the derivative of numerator over derivative of the denominator.
May 24, 2017

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

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#