How do you find the limit of #f(x) = (x^2 + x - 6) / (x + 3) # as x approaches 0?

1 Answer
Feb 18, 2016

-2

Explanation:

#lim_(x rarr 0) (x^2+x-6)/(x+3)=(0^2+0-6)/(0+3)=-6/3=-2#

There is no indetermination. Maybe you have made a mistake.

Did you mean

#lim_(x rarr -3) (x^2+x-6)/(x+3)#

that in fact gives #0/0#?

Calculate the division of #x^2+x-6# by #x+3# which gives #x-2#.

#lim_(x rarr -3) (x^2+x-6)/(x+3)=lim_(x rarr -3)((x-2)cancel((x+3)))/cancel(x+3)=lim_(x rarr -3) (x-2)=-3-2=-5#