Question

How do you find the limit of `( sqrt(x^2+11) - 6 ) / (x^2 - 25)` as x approaches 5?

Answer 1

`1/12`

Explanation:

Notice how evaluating the limit presently would be impossible, since the denominator of the fraction would be `0`.

A good strategy when trying to deal with expression with radicals and other general nastiness is to multiply by a conjugate. We can try to multiply this by the conjugate of the numerator.

`=lim_(xrarr5)(sqrt(x^2+11)-6)/(x^2-25)((sqrt(x^2+11)+6)/(sqrt(x^2+11)+6))`

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`.

`=lim_(xrarr5)((sqrt(x^2+11))^2-6^2)/((x^2-25)(sqrt(x^2+11)+6))`

`=lim_(xrarr5)(x^2+11-36)/((x^2-25)(sqrt(x^2+11)+6))`

`=lim_(xrarr5)(x^2-25)/((x^2-25)(sqrt(x^2+11)+6))`

Notice the `x^2-25` terms in the numerator and denominator will cancel.

`=lim_(xrarr5)1/(sqrt(x^2+11)+6)`

We can now evaluate the limit by plugging in `5` for `x`.

`=1/(sqrt(25+11)+6)=1/(sqrt36+6)=1/12`

We can check a graph of the original function:

graph{( sqrt(x^2+11) - 6 ) / (x^2 - 25) [-1.53, 12.52, -0.05, .15]}

Even though the point at `x=5` is undefined, it is very close to `1/12=0.08bar3`.