How can finding limits by conjugates be used to solve for rational expressions?

[atomicpedals]

The other day in a fit of boredom I decided to dust off my old math books (high school and undergrad) and see if I can still do basic calculus. These days if I need to solve anything I ask a computer to do it, the hazards of getting a job in industry I suppose.

All that said, I have been tripped up by finding limits by conjugates.

1. Homework Statement

Rationalize the following expression by conjugates: $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2}$$

Homework Equations

Not applicable.

The Attempt at a Solution

So the easy way to solve this is to simply plot the function, there you clearly see that the limit is 4. However, that's too easy as this can clearly be done by hand (otherwise the exercise would have said "graph it"). The basic algebra is where I go horribly astray.

$$ \frac { (x+1) } { \sqrt{x+5} - 2} = \frac { (x+1) } { \sqrt{x+5} - 2} \frac { \sqrt{x+5} + 2 } {\sqrt{x+5} + 2} $$
Having cheated and graphed the function I can tell by inspection that the denominator should go to $(x + 1)$ to cancel out the $(x + 1)$ in the numerator and yield the limit of 4. And as I'm writing this I think I've answered my own question (huzzah!); it's a simple application of FOIL. $$ ( \sqrt {x + 5} - 2 ) ( \sqrt {x + 5} + 2 ) = x + 5 - 4 = (x + 1) $$ And so $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2} = \lim_{x \rightarrow -1} \sqrt{x+5} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 $$ Have I gone about this the right way or did I just get lucky?

 

[Dick]

Well done!

 

[fresh_42]

Have I gone about this the right way or did I just get lucky?

Both.

This is a nice example how inserting the value for $x$ does not work: instead of $\dfrac{0}{0} =$ indeterminate, we get a real limit. It also shows why division by zero is prohibited: it could be any value and hence no meaningful one.

Additionally it shows, that the function $x \longmapsto \dfrac{x+1}{\sqrt{x+5}-2}$ can be continuously extended, although it is not defined at $x=-1$.