How do you find the limit at a hole?

The function I have is #lim_(x rarr 9) (x-9)/(sqrt(x+7)-4.)# How do I find the limit, if the function gives #0/0?#

1 Answer
Apr 5, 2018

#lim_(x->9) (x-9)/(sqrt(x+7)-4) = 8#

Explanation:

We can find this limit algebraically by eliminating a common factor that is causing the hole. It is slightly cleaner to see if we put #t = x+7# as follows...

#lim_(x->9) (x-9)/(sqrt(x+7)-4) = lim_(t->16) (t-16)/(sqrt(t)-4)#

#color(white)(lim_(x->9) (x-9)/(sqrt(x+7)-4)) = lim_(t->16) (color(red)(cancel(color(black)((sqrt(t)-4))))(sqrt(t)+4))/color(red)(cancel(color(black)((sqrt(t)-4))))#

#color(white)(lim_(x->9) (x-9)/(sqrt(x+7)-4)) = lim_(t->16) sqrt(t)+4#

#color(white)(lim_(x->9) (x-9)/(sqrt(x+7)-4)) = sqrt(16)+4#

#color(white)(lim_(x->9) (x-9)/(sqrt(x+7)-4)) = 8#