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

Beginner2023-03-07Added 3 answers

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 $(a+b)(a-b)={a}^{2}-{b}^{2}$.

$=\underset{x\to 5}{lim}\frac{{\left(\sqrt{{x}^{2}+11}\right)}^{2}-{6}^{2}}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

$=\underset{x\to 5}{lim}\frac{{x}^{2}+11-36}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

$=\underset{x\to 5}{lim}\frac{{x}^{2}-25}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

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 $\frac{1}{12}=0.08\overline{3}$.

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 $(a+b)(a-b)={a}^{2}-{b}^{2}$.

$=\underset{x\to 5}{lim}\frac{{\left(\sqrt{{x}^{2}+11}\right)}^{2}-{6}^{2}}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

$=\underset{x\to 5}{lim}\frac{{x}^{2}+11-36}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

$=\underset{x\to 5}{lim}\frac{{x}^{2}-25}{({x}^{2}-25)(\sqrt{{x}^{2}+11}+6)}$

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 $\frac{1}{12}=0.08\overline{3}$.