# How do you show the limit does not exist lim_(x->4)(x-4)/(x^2-8x+16) ?

Aug 4, 2014

=$\infty$

Solution

=lim_(x→4)((x-4)/(x^2-8x+16)), plugging the limit we get $\frac{0}{0}$

=lim_(x→4)((x-4)/(x^2-4x-4x+16))

=lim_(x→4)((x-4)/(x(x-4)-4(x-4)))

=lim_(x→4)(x-4)/((x-4)^2)

=lim_(x→4)1/((x-4))

Now applying the limit, we get

$= \frac{1}{0} = \infty$, which implies limit does not exist