How can I solve limit problems? I've never been able to understand written or verbal explanations of math concepts. The only way I can solve math problems is by seeing the process done step-by-step on similar problems and memorizing the patterns.
If someone were to ask me, "What's 3 divided by 4?" I wouldn't be sure if that means "3/4" or "4/3." The only way I'd know the answer is to refer to the similar problem of "What's 100 divided by 4," and I've memorized that it means "100/4."
I'm not sure if I'm capable of doing calculus. How can I solve these problems?
http://imgur.com/Hjp6A9c
I asked the teacher, and he replied with:
"Here is the solution for part (c). Part (b) would be similar. Each example has to be done separately so that is why the answers change around.
The crucial points are x = -2, 1, 3. We can get the sign of f(x) between and beyond these numbers. (If the sign of f(x) is going to change it will be at these numbers.)
x < -2: x - 3 < 0, x + 2 < 0, x - 1 < 0 so f(x) = - / - - = -/+ = - so f(x) < 0.
-2 < x < 1: x - 3 < 0, x + 2 > 0, x - 1 < 0 so f(x) = - / + - = -/- = + so f(x) >0
1 < x < 3: x - 3 < 0, x + 2 > 0, x - 1 > 0 so f(x) = -/ + + = -/+ = - so f(x) < 0
x > 3: x - 3 > 0, x + 2 > 0, x - 1 > 0 so f(x) = +/ + + = +/+ = + so f(x) > 0
Thus for x --> 1-, -2 < x < 1 so f(x) > 0 thus f(x) --> + infinity. For x --> 1 +, 1 < x < 3 so f(x) < 0 thus f(x) --> - infinity.
x < -2: For the sub-interval of the domain where x is less than -2
x - 3 < 0, x + 2 < 0, x - 1 < 0 The factors in the rational expression have these signs
so f(x) = - / - - = -/+ = - So that the rational expression in this sub-interval has a negative factor on top being divided by two negative factors. The two negative factors give a positive on multiplication so overall it is a negative divided by a positive,
so f(x) < 0. which means the rational expression is negative in this sub interval. Thus as y goes towards the asymptote at -2 from below it must be going down so y {= f(x)} goes to minus infinity.
The asymptotes are at values of x which make the denominator zero. The asymptotes are where the function will go to infinity. That is why the question asks you to start with finding those values for x. To find exactly where y goes as x approaches those values you need to know the sign of the rational expression in the intervals of the domain on either side of the asymptote. That tells you if it is plus or minus infinity that y goes to. To find the sign in the intervals you have to find the signs of each of the terms then using sign number arithmetic rules you can get the sign of the expression."
I don't understand any of that. The only thing I see is that, for C, x=-2 and y=1.
I don't understand why the positive and negative switch between problems B and C. The problems have similar set-ups, so I don't see why the solution for the last y in B is positive infinity and the solution for the last y in C is negative infinity.
What do the little plus and minus signs mean? What does "-/--" mean? What do the "-/+" symbols mean? How are the "<" and ">" symbols used to solve the problem? I don't see them in the problem itself.
How would I find the signs of each of the terms, in order to see if the infinity would have a plus or minus sign?
 A: All of the little $+$ and $-$ signs represent the signs of the various factors in various locations, and symbols such as "$-/--$" are meant to suggest how the signs combine in the overall expression (here, it means a negative factor in the numerator, with two negative factors in the denominator).
The idea is easy to understand. I'm sure you already know how to multiply signs (division works exactly the same way, so I won't distinguish between them). For example, you know that $(-237)(-129)$ is a positive number before you actually multiply it out to get an exact answer, right? The important thing is that you see two negative factors, which results in a positive product. Same thing with $(\frac{-129}{13})({-77751})$; you can answer immediately that the product is a positive number before you work on getting the exact answer. The idea in these cases can be abbreviated as "(-)(-)=(+)", short for "a negative times a negative is a positive", or even more compactly as "$--=+$". That's all this symbol means. Same with division: "a negative divided by a negative is a positive", so "$(-)/(-)=(+)$" or, more compactly, "$-/-=+$" (specific example: $(-6)/(-2)=3$; of course $3$ is $+3$, but we usually suppress the explicit $+$ sign on positive numbers even though it is really understood to be present).
To find the signs of each of the factors, deal with the factors one at a time. Assuming they are linear factors (like $x-17$ or $2x-1$ or $\frac12x+\frac23$), then for a given factor, first find where that factor is zero. Why? Because it will be zero in exactly on place, so it will have a constant sign on one side of that place, and a constant sign on the other side of that place. It can only change sign by passing through zero, so the place where the factor is zero breaks the number line into two halves with a constant sign for that factor on each side.
That's where the symbols $<$ and $>$ come in. Suppose we are considering the factor $x-3$. As above, we first find where it is zero: where is $x-3=0$? Well, solve that equation for $x$ to get $x=3$. We now know that the factor $x-3$ is zero only at $x=3$. But what about everywhere else? What about values of $x$ that are not $3$? If $x$ is not $3$, then it can be either to the right of $3$ or to the left of $3$; in other words, it can be either greater than $3$ (written $x>3$) or less than $3$ (written  $x<3$). Now you can tell the sign of the factor on each side, because on the right side, we have $x>3$, so (subtracting $3$ from both sides of this inequality) $x-3>3-3$ which just says $x-3>0$ and so the factor is positive. 
Likewise on the left side for this example we have $x<3$ which is the same as $x-3<0$ which says the factor is negative.
Note that you could just pick any number on each side and plug it into the factor to see if the result is positive or negative, and that will tell you the sign for ALL choices on that side (because the sign is constant on each side, so it doesn't matter where you sample the side).
Finally, you can align your sign charts for each factor and combine signs multiplicatively as I first discussed. Remember when doing this that if a factor is zero in the numerator, the product will be zero at that point; and if a factor is zero in the denominator, the result is undefined (infinite? Not a number at any rate) at that point.
Let's put your (c) together in a combined sign chart:
$$\begin{array} \textrm{line}: & \cdots& -2 & \cdots& 1 & \cdots& 3 & \cdots\\
x-3: &-&-&-&-&-&0&+\\
x+2: &-&0&+&+&+&+&+\\
x-1: &-&-&-&0&+&+&+\\
\tfrac{x-3}{(x+2)(x-1)}: &-&?&+&?&-&0&+\\
\end{array}$$
The top row shows the various intervals on the number line, which result from the  points of interest where any of the factors is zero.
The next three rows show the sign charts for each factor separately.
The las row combines the factor signs into a sign chart for the whole expression, remembering to handle multiplication or division by zero specially.
I hope this helps.
