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

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Answer 1 · 2017-03-18T22:06:07

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

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]}

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