TITLE: Finding limit of a quotient
QUESTION [6 upvotes]: I can't find this for some reason. I know I asked about 6 of these before, and I was able to finish my homework but now I went back to review and I can't do a single one of these problems on my own. Even the ones I did figure out on my own. I spent probably a total of 14 hours on the homework, not sure what is wrong with me but I don't know how to factor or do basic algebra. 
Anyways I need to find 
$$\lim_{x\to -2}\frac{x+2}{x^3+8}.$$
I have spent at least an hour on it and I can't figure it out.

REPLY [5 votes]: A clever substitution might even gift you the answer. 
It is usually easy to take the limit of a function about some special points, like $0$, $1$, $\pm \infty$ etc, since it is easier to get a feel for functions around such points. Also, most of the elementary limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x-1}{x} = 1$ already have the arguments approaching $0$ anyway, so it may be convenient if a new complicated limit looks like this as well. 
Finally, as in any calculation, you should be especially careful while dealing with negative quantities. In your question, for instance, the negative sign in $-2$ could lead to some confusion while factoring, unless you are systematic in your long division or synthetic division.  
In the light of the above points, I recommend the substitution strategy for evaluating limits. That is, if a given limit has $x \to a$, make the substitution $y = x-a$ and let $y \to 0$. Plug in $x=y+a$ in the function, and see if the resulting expression is simpler. As a rule of thumb, I would expect this trick would be powerful when you are dealing with polynomial or rational functions, or even exponential function $e^x$. On the other hand, if your limit involves trigonometric functions, then I would guess that the resulting expression would become more complicated, basically bacause the sum formulas of trigonometric functions involve multiple terms. 
Let's see what this does in your example. Let $y = x+2$, so that $$x^3+8 = (y-2)^3+8 = y^3 - 8 - 3\cdot y \cdot 2(y-2) +8 = y^3 - 6y^2 + 12 y .$$ Hence, the given limit becomes:
$$
\lim_{y \to 0} \frac{y}{y^3 - 6y^2 + 12y} = \lim_{y \to 0} \frac{1}{y^2 - 6y + 12} = \frac{1}{12} .
$$
Notice that the evaluation of the limit in the final step is rather obvious. 
Of course, I must warn you that you cannot blindly apply this (or, rather any) trick in all cases. For instance, given the limit
$$
\lim_{x \to \pi/2} \frac{\exp(\cos x) - 1}{\cos x},
$$
rewriting it as 
$$
\lim_{y \to 0} \frac{\exp(\cos (y + \pi/2)) - 1}{\cos (y+\pi/2)}
$$
seems hardly useful. (Of course, substitution is the way to go in this example, just
 not this one.)