How to find the limit of x 2 + 11 - 6 x 2 -...

Answer: ${\displaystyle \frac{1}{12}}$
Consider how evaluating the limit now would be impossible because the fraction's denominator would be 0.
When dealing with radical expressions and other general nastiness, multiplying by a conjugate is a good strategy. We can try multiplying this by the numerator's conjugate.
${\displaystyle =\underset{x\to 5}{lim}\frac{\sqrt{x^2+11}-6}{x^2-25}\left(\frac{\sqrt{x^2+11}+6}{\sqrt{x^2+11}+6}\right)}$
In the numerator, we can recognize that this is really the difference of squares pattern in reverse, where ${\displaystyle (a+b)(a-b)=a^2-b^2}$.
${\displaystyle =\underset{x\to 5}{lim}\frac{{\left(\sqrt{x^2+11}\right)}^2-6^2}{(x^2-25)(\sqrt{x^2+11}+6)}}$
${\displaystyle =\underset{x\to 5}{lim}\frac{x^2+11-36}{(x^2-25)(\sqrt{x^2+11}+6)}}$
${\displaystyle =\underset{x\to 5}{lim}\frac{x^2-25}{(x^2-25)(\sqrt{x^2+11}+6)}}$
Notice the ${\displaystyle x^2-25}$ terms in the numerator and denominator will cancel.
${\displaystyle =\underset{x\to 5}{lim}\frac{1}{\sqrt{x^2+11}+6}}$
We can now calculate the limit by substituting 5 for x.
${\displaystyle =\frac{1}{\sqrt{25+11}+6}=\frac{1}{\sqrt{36}+6}=\frac{1}{12}}$
Even though the point at ${\displaystyle x=5}$ is undefined, it is very close to ${\displaystyle \frac{1}{12}=0.08\overline{3}}$.