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The function I'm looking at is this: $f(x)=\frac{\sqrt {x^2+2x+6}-3}{x-1}$

I'm not entirely sure when to use rationalization vs. taking values close to $x=a$ when finding limits. I realize I can make $x$ equal values close to $1.$ I can also rationalize the function to $f(x)=\frac{x-1}{\sqrt{x^2+2x+6}+3}$ by multiplying the denominator and numerator by the conjugate of the numerator and use $x=1$ exactly.

When do I use each method? Thank you!

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    What are you asking? Rationalization simply reduces the expression to a form that facilitates evaluation for values close to "$a$." – 2017-02-27
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    Rationalization is more "rigorous" than trying values. On the other hand, if you can use LHopital that is valid also and often easier. – 2017-02-27
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    Thank you. What I'm trying to clarify in my own mind: Is one method used for finding $\lim_{x\to a}$, and the other method for finding $\lim_{x\to a^-}$ and/or $\lim_{x\to a^+}$? – 2017-02-27

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Well, the answer is that it depends on the specific problem. If $f(x)$ is continuous at $x=a$, then $\lim_{x\to a}f(x)=f(a)$ and we can simply "plug" in the value $a$.

For forms such as the one herein, $f(x)$ actually has a removable singularity at $x=1$ (but we don't know this a priori). If we "plug in values close to $1$," then we get a form that appears superficially to be of indeterminate form.

We could apply L'Hospital's Rule here and proceed without any difficult. We could also use Taylor's Theorem and expand the numerator around $1$. In the latter case, we would find that the zeroth order term in the expansion is $0$, and that the first order term in non zero. Hence we expose the removable discontinuity.

The most elementary approach is to rationalize. Rationalization simply reduces the expression to a form that facilitates evaluation of the limit. Note that we can write

$$\begin{align} f(x)&=\frac{\sqrt{x^2+2x+6}-3}{x-1}\\\\ &=\left(\frac{\sqrt{x^2+2x+6}-3}{x-1}\right)\left(\frac{\sqrt{x^2+2x+6}+3}{\sqrt{x^2+2x+6}+3}\right)\\\\ &=\frac{x^2+2x-3}{\left(\sqrt{x^2+2x+6}+3\right)(x-1)}\\\\ &=\frac{\color{blue}{(x-1)}(x+3)}{\left(\sqrt{x^2+2x+6}+3\right)\color{blue}{(x-1)}}\\\\ &=\frac{x+3}{\sqrt{x^2+2x+6}+3} \end{align}$$

And now, we can plug and finish!

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    Thank you for your answer. My question isn't how to rationalize. My question is when. When do I rationalize vs. when do I plug in "close to" values? – 2017-02-28
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    I've edited to add some thoughts around your very good question. -Mark – 2017-02-28
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    Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – 2017-03-06
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In a nutshell: plugging in values is a way to get an idea of how the function behaves near $a$, it is not a method of computing the limit. Rationalization is one of many methods.

My question isn't how to rationalize. My question is when. When do I rationalize vs. when do I plug in "close to" values?

When do I use each method? Thank you!

It depends: what do you want to do? They serve different purposes. If you want to get an idea of how the function behaves near $a$, you can try plugging in values close to $a$. This will however never be a rigourous argument to show that the limit exists and if it exists, what its value is. It will usually give you a good indication about the existence of the limit, or if it tends to $\pm \infty$ for example.

If you want to find the limit and prove it exists (or not), simply plugging in some values is insufficient. Rationalization is one of possibly many methods to find the limit, but there are other ones: l'Hôpital's rule and the use of Taylor series were already mentioned, there is of course also the formal definition you could use, et cetera.

Which method to choose depends on what you have seen and what you are allowed to use. If you have multiple options, it can heavily depend on the function which method is most appropriate. In your example, rationalization is definitely a good choice but other approaches can work as well.