Question #55f2d

2 Answers
Sep 9, 2016

#lim_(xrarr2) (x^3-3x^2+4)/(x^3-x^2-8x+12) = color(green)(3/5)#

Explanation:

The problem is that both:
#color(white)("XXX")x^3-3x^2+4# and
#color(white)("XXX")x^3-x^2-8x+12#
are equal to #0# if #x=2#

...but this means that both of these expressions are divisible by #(x-2)#

and using either synthetic or long division we can get:
#color(white)("XXX")(x^3-3x^2+4)/(x^3-x^2-8x+12)=(cancel(""(x-2))(x^2-x-2))/(cancel(""(x-2))(x^2+x-6)#

Unfortunately both
#color(white)("XXX")x^2-x-2# and
#color(white)("XXX")x^2+x-6#
are also both equal to #0# if #x=2#

...but (again) this means that both of these expressions are divisible by #(x-2)#

and we can get
#color(white)("XXX")(x^2-x-2)/(x^2+x-6)=(cancel(""(x-2))(x+1))/(cancel(""(x-2))(x+3))#

So we have
#color(white)("XXX")lim_(xrarr2) (x^3-3x^2+4)/(x^3-x^2-8x+12) =lim_(xrarr2) (x+1)/(x+3)#

#color(white)("XXXXXXXXXXXXXXXXX")=(2+1)/(2+3)=3/5#

Sep 9, 2016

I found: #3/5#

Explanation:

Try this:
enter image source here