

Limits and all that silly stuff
2  The format of a limiting equation is "lim x > a y = equation". This is read as: As x approaches the constant 'a', what does the equation go to?" In the problem, there are two limits involved, and throughout the section you will have look at both of them. There is a limit of x, and a limit of y (which is limit of equation). The problem explicitly states where x goes to. As x approaches that number, the whole equation will approach another. That is the other limit; that is where y goes to. y is equal to the whole equation. I will now start with basic problems. Grasp each before you move on. 3  Limits of x are on the horizontal axis, and limits of y are on the
vertical axis. This should be obvious, but everyone mixes the two up some time
or another. Especially with asymptotes. You will learn about asymptotes soon. Simple LimitsA limit says that x is approaching a number, ranging from negative infinity to positive infinity. Lim x>a y = 3x + 4 y = 3a + 4 In that case, x was going to a, so you plug a into the equation. Simple enough. It comes out the same as x = a. Lim x>0 y = 3x + 4 y = 3(0) + 4 = 4 That equals 4, because the other term is approaching 0. Lim x>∞ y = 3x + 4 y = 3(∞) + 4 = ∞ + 4 = ∞ This is where it begins to get more complicated. As the first term approaches infinity, so does the entire function. Even bigger than infinity, technically, (infinity plus four) but that is infinity. Infinity is not a specific number, so anything that just keeps on getting larger without end is said to be going to infinity. Note that when we say the equation is going to a number, or infinity, it is referring to the yaxis of the equation, as x approaches the said number. In this example, both x and y go to infinity. In the following picture, however badly drawn, as x approaches 0 from the positive side, the yaxis approaches negative infinity. This is an example of how when we say that this function is going to negative infinity we are referring to the yaxis. If you look at the y value at x = 1/2, you will notice it is far less than x = 1, and x = 1/10 has a y value far less then f(1/2). Incidentally, this picture is for function y = ln x, and it will be discussed below. For more on natural logarithms, you can go to my precal page.
Lim x>3 y = (x^{2} + x 12)/(x 3) This is not simple. If you plug in 3 then the equation is undefined, because it goes to 0/0. What you can do however is factor out the x3. In problems like this you can usually factor out. It becomes: Lim x>3 y = (x 3)(x + 4)/(x 3) y = x + 4 y = 7 I canceled out the (x  3) and it turns out that as x approaches 3, the yaxis of the equation goes to 7. Notice however that it NEVER equals seven! It gets infinitely close. There is a gap in the equation at the coordinates (3,7).
Dealing with x going to infinityWhen x goes to infinity, you must look to see if x is in the numerator (or if it's not a fraction, same thing), if x is in the denominator, or both. I will cover these three scenarios. If x is either in the numerator (top) of a fraction or in a regular number, the curve is going to go higher on the yaxis as x approaches infinity. An example is the simple equation y = x^{2} As x gets bigger, so does y. Another scenario is where x is on the bottom of a fraction. A simple case is y = 1/x. Here, as x gets very big, the equation becomes very small. (or say that y gets small) When x is a million, the equation equals one millionth, an extremely small number. So in this case, the limit as x goes to infinity is 0, because it is approaching that. It will never actually be zero, but that doesnt matter in a limit against infinity. Regardless, infinity is impossible, since it does not imply reality, but in the world of infinity, this equation actually does equal 0. This is a horizontal asymptote. In any scenario where the bottom of the fraction approaches ∞, the equation goes to 0. In any scenario where the top approaches ∞, the equation goes to ∞. Lim x>∞ y = 48/x^{2} is 0 Lim x>∞ y = 48x^{2}/x = 48x is ∞ Now that you have a simple understanding of what is happening, I will propose a general rule for the next examples to deal with every limit where x goes to infinity. If you try it for cases where x does not go to infinity, you will be totally screwed up, and I take no blame. Lim x>∞ y = (3x^{2} + 5x)/(7x^{3} 2) In any case where x is going to infinity involving a
fraction, you must only care about the biggest factor of x in the equation. Every
other term goes to 0. So this equation can be retyped as (0 + 0)/(7x^{3}
0). Needless to say, this equation goes to 0. This is because the biggest
factor is on the bottom. Lim
x>∞
y = (5x^{3} 2x^{2} + 17)/(8x^{2} 4) =
5x^{3}/0 = ∞ That was a case where
the biggest factor was on top. What happens if there is a biggest factor on both
the top and bottom? Lim
x>∞
y = (x^{2} 1)/(4x^{2} + 6x) = x^{2}/4x^{2}
= 1/4 That case has the
biggest factor of x, x^{2}, on both the top and bottom. But once we
rewrite the equation, using my standard rule of largest factor of x, they will just cancel out, leaving behind their
coefficients! Lim
x>∞
y = (x^{2} 1)/(4x^{2} + 6x) = x^{2}/4^{2}
= 1/4 There are still more complex scenarios, including the ones where x goes to negative infinity, or the problem is set up so that the function (or y) will end up going to negative infinity. In the following case it turns out that the equation goes to negative infinity. The reason is that using the previous methods, the only factor left is 5x^{3}, and whenever the factor going to infinity is negative, it goes to negative infinity. (Note that if you are dealing with a fraction, and there is a remaining factor on the top and bottom with negative, they will cancel out, like 1's) Lim x>∞ y = (5x^{3} 2x^{2} + 17)/(8x^{2} 4) = 5x^{3}/0 = ∞ In the following case you must realize that negative infinity squared is positive, and cubed is negative again, like any negative number. Do not confuse it with the past case where there is a negative coefficient! Lim x>∞ y = x^{2} = (∞)^{2} = ∞ Lim x>∞ y = x^{3} = (∞)^{3} = ∞
One sided
limits
If I say lim x> 0^{} then I mean if x is approaching 0 from the left, on a graph. If I say lim x > 0^{+} then I mean x is approaching specifically from the right. This is important in some cases. Lim x>0^{+} y = ln x If you know what the graph looks like then you know the answer is ∞. The point here is that I had to say from the right because on the graph of ln x there is no way for x to be less than 0, so it cant be approaching from there. It is clear from the picture, but you can try to solve ln 2 on a calculator; you will get an error. Lim x>0^{} y = 3/x = ∞. If I had said 0^{+} then the equation would be big on the positive side. Look at the graph 1/x, below. It is comparable to 3/x. It does approach negative infinity as the graph gets close to 0, and once you cross it, it comes down from positive infinity.
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