11. $lim_{x\rightarrow6}left(\frac{x + 2}{x^{2}-4x - 12}\right)=$

Explanation:

Step1: Factor the denominator

First, factor $x^{2}-4x - 12=(x - 6)(x+2)$. So the limit becomes $\lim_{x\rightarrow6}\frac{x + 2}{(x - 6)(x + 2)}$.

Step2: Simplify the function

Cancel out the common factor $(x + 2)$ (for $x\neq - 2$ and as $x\rightarrow6$). We get $\lim_{x\rightarrow6}\frac{1}{x - 6}$.

Step3: Analyze the limit

As $x\rightarrow6$, the denominator $x - 6\rightarrow0$. The numerator is a non - zero constant 1. When $x$ approaches 6 from the right ($x>6$), $x-6>0$ and $\frac{1}{x - 6}\rightarrow+\infty$. When $x$ approaches 6 from the left ($x<6$), $x - 6<0$ and $\frac{1}{x - 6}\rightarrow-\infty$. So the limit does not exist.

Answer:

The limit does not exist.