How do you evaluate the limit #lim((x-3)^2-9)/(2x)dx# as #x->0#?

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Answer 1 · 2016-11-20T01:19:28

The limit equals #-3#.

First expand.

#=>lim_(x->0) (x^2 - 6x + 9 - 9)/(2x)#

#=> lim_(x->0) (x^2 - 6x)/(2x)#

#=>lim_(x->0) (x(x - 6))/(2x)#

#=>lim_(x->0) (x- 6)/2#

#=>(0 - 6)/2#

#=> -3#

Hopefully this helps!