You can solve this problem through factoring.
First, you have the equation,
[tex]h(x) = \frac{x^2-36}{x-6}[/tex]
Then, you can factor the numerator.
[tex]h(x) = \frac{(x+6)(x-6)}{x-6}[/tex]
You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with [tex]h(x) = x+6[/tex]
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
