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#