How do you find the limit of #(x^2 - 8) / (8x-16)# as x approaches #sqrt8^+#?

1 Answer
Mar 18, 2017

#lim x->sqrt(8)^+ = 0#

Explanation:

Remember that a limit value is a #y#-value.

Just plug in #x = sqrt(8)# into the function:

#((sqrt(8))^2-8)/(8sqrt(8)-16) = (8-8)/(8sqrt(4)sqrt(2)-16) = 0/(16sqrt(2)-16) = 0#

From the graph you can see that at #sqrt(8) ~~ 2.8284# the function crosses the #y=0 # line #(x-"axis"):#
graph{(x^2-8)/(8x-16) [-10, 10, -5, 5]}