How do you evaluate the limit #((3+h)^3-27)/h# as h approaches #0#?

1 Answer
mason m
Dec 5, 2016

#lim_(hrarr0)((3+h)^3-27)/h=27#

Explanation:

The first step is to simplify #(3+h)^3=(h+3)^3#, either by multiplying out or by using the binomial theorem if you know it.

#(h+3)^3=(h+3)^2(h+3)#

#=(h^2+6h+9)(h+3)#

#=h^3+9h^2+27h+27#

Then:

#lim_(hrarr0)((3+h)^3-27)/h=lim_(hrarr0)((h^3+9h^2+27h+27)-27)/h#

#=lim_(hrarr0)(h^3+9h^2+27h)/h#

#=lim_(hrarr0)(h(h^2+9h+27))/h#

#=lim_(hrarr0)(h^2+9h+27)#

#=0^2+9(0)+27#

#=27#

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