Jamya White

2023-03-06

How to find the limit of $\frac{\sqrt{{x}^{2}+11}-6}{{x}^{2}-25}$ as x approaches 5?

nanaandhachi9ur

Answer: $\frac{1}{12}$
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.
$=\underset{x\to 5}{lim}\frac{\sqrt{{x}^{2}+11}-6}{{x}^{2}-25}\left(\frac{\sqrt{{x}^{2}+11}+6}{\sqrt{{x}^{2}+11}+6}\right)$
In the numerator, we can recognize that this is really the difference of squares pattern in reverse, where $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$.
$=\underset{x\to 5}{lim}\frac{{\left(\sqrt{{x}^{2}+11}\right)}^{2}-{6}^{2}}{\left({x}^{2}-25\right)\left(\sqrt{{x}^{2}+11}+6\right)}$
$=\underset{x\to 5}{lim}\frac{{x}^{2}+11-36}{\left({x}^{2}-25\right)\left(\sqrt{{x}^{2}+11}+6\right)}$
$=\underset{x\to 5}{lim}\frac{{x}^{2}-25}{\left({x}^{2}-25\right)\left(\sqrt{{x}^{2}+11}+6\right)}$
Notice the ${x}^{2}-25$ terms in the numerator and denominator will cancel.
$=\underset{x\to 5}{lim}\frac{1}{\sqrt{{x}^{2}+11}+6}$
We can now calculate the limit by substituting 5 for x.
$=\frac{1}{\sqrt{25+11}+6}=\frac{1}{\sqrt{36}+6}=\frac{1}{12}$
Even though the point at $x=5$ is undefined, it is very close to $\frac{1}{12}=0.08\overline{3}$.

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