 Now that you know your limit laws, remember the laws only when they do exist, when the limit actually exists you can apply them, now let's calculate some limits, we've got a lot of knowledge now about limits, let's put that knowledge to work, let's calculate some limits, now we've seen the laws of limits but now we've actually got to develop a few recipes to solve these problems, let's start with the very first one. We have the limit as T approaches 2 of T squared minus 4 over 2 minus T, well clearly I cannot use the limit laws here, because if I were to take the limit as T approaches 2 of the numerator, certainly I'll get to zero, but if I take the limit of T approaches 2 of the denominator that's also going to equal zero, not divide by zero, and therefore as it stands this limit doesn't exist, it actually does, but as it stands it doesn't and you cannot use the limit laws here. Look at what I've done in step one here, I have noticed though that the numerator can be factorized into T minus 2 and T plus 2, that's beautiful, T squared minus 4 it's been factorized there. What I also noticed in the denominator is if I take a negative 1 out of this denominator, that polynomial of first degree in the denominator, I get this, negative T minus 2, look if I distribute the negative in there I'm still going to get 2 minus T and now I noticed I can cancel these two out, they can cancel leaving me with a negative 1 and the T plus 2, I can make use of a limit law now by taking this constant in negative 1 outside of the limit, if that's outside of the limit there we are, now I just have the limit as T approaches 2 of this T plus 2, I can now just put the 2 in there and I have a negative 4, beautiful by making use of factorization of a polynomial, I can solve this problem, this is the first proper way other than just plugging in values closer and closer to as we did in the tables way long time ago, this is the first proper way of doing it. I'm just factorizing the numerator, I can cancel common factors out here, make use of my limit laws and get to a value of negative 4. Now here's a second method and we actually just using our limit laws, this is a beautiful example, you'll note here that I can't factorize, I can't factorize at all. But what I would notice is that this actually just says the limit as T approaches 2 of some function, the f of T and in this instance says the f of T, but this f of T is to the power 1 over 3, that's what this cube root is, it's to the power 1 over 3. Which means I can actually do this or this, one of my limit laws, I can just take this power outside of the limit so I can do the limit, first do the limit of my function f of T and then take it to a power third or the cube root day. Now this is quite an easy limit to do, I can simply plug in 2, 2 squared is 4 times 3 is 12 minus 4 is 8, so it's just a cube root of 8 and I remember 8 is 2 cubed to the power 1 over 3, well that's just 2. So in the end what might seem a bit difficult to do is actually quite easy, I needn't even have taken out this cube root as I've done here, I might as well just have plugged the 2 in and I would have gotten the cube root of 8 which is 2 and simply got to the answer there. So quite a few ways of doing this, but here we made use of the fact that we can take this power here which is actually just to the power third, so it is this function to the power third, I can take this outside of my limit day, excellent. Now here's an exciting example, quite a difficult problem if we just look at it now, now we've just seen how to do factorization. We've seen what to do just with a cube root or just powers I should say, and now we've got this very complicated example. First of all if we were just to substitute the 0 in there we'll have 2 squared here minus 2 squared there again 0 divided by a 0 there. We cannot have this, this is not going to work for us. Let's try something else. One thing that we could do is we can multiply this fraction out by something else. If I were to multiply it by 2 over 2, 2 over 2 is just 1, so I'm not changing this fraction of polynomials at all, but I'm not going to multiply it by 2 over 2, I'm going to multiply it with this. Where there was a negative sign here we're going to have a positive sign there, so it's the square root of t plus 2 minus the square root of 2. I'm going to have the square root of t plus 2 here plus the square root of 2, so I know if I do that I have this factorization issue. In other words if I were to multiply these two out by each other this is what I'm going to be left with. Do that, it's quite a few steps but it's quite simple algebra. I'm going to be left with this. Now what I've done in the numerator I've also got to do here in the denominator. And let's just have that, I'm just going to write the t out, so I've just got the t multiplied by this new factor in the denominator there. So the numerator becomes quite easy in that I have t plus 2 minus 2, so I'm just left with t and in the denominator I still have my t there and these two can now cancel each other out, leaving me with 1 over this factor in the denominator. If I were to just plug in 0 into this I'm not going to be left with this 2 minus 2 problem that I have at the top, but actually a plus, a positive at the bottom. So I'm going to be left with the square root of 2 plus the square root of 2 which is twice the square root of 2. So quite a simple answer. So always remember this little bag of tricks or keep it in your bag of tricks this multiplying out one of the two, either the numerator or denominator by something which would then make it into the factor of something which is a lot easier. So so far in our bag of tricks we've seen direct substitution. I just want to go over this again, you can just directly substitute whatever this limit is that we try to approach into your expression. And if that works out fine that's fantastic. Number two is we could factorize and number three we rationalize the numerator. That's what we did in the last example actually we rationalize the numerator. So those are the three things we can do at this time remembering our limit laws which just make things a little bit easier for us. Let's do this example. Which one of those methods we have in our bag of tricks do you think we should try here? Quite clearly we are going to just take what just decompose this into factors. T squared minus 2 T minus 8 can be rewritten as this T plus 2 and T minus 4. And the T squared minus 3 T minus 4, well that's quite easy to deconstruct that's a T minus 4 and a T plus 1. Now why did we not use the direct substitution method? We'll try that initially and what are you going to get? You're going to get zero over zero. Now that is something very special and in later chapters we will deal with that. But now look at this we have a T minus 4 there and a T minus 4 there. Now we just left with T plus 2, T plus 2 in the numerator and T plus 1 in the denominator. If we were now to substitute 4 directly into this it is quite a simple fraction. Another example we have the limit as T approaches 2 of T cubed minus T minus 6. First things first always try direct substitution. You might get an easy problem like this. If we were just to substitute 2 in we'll have 8 minus 2 minus 6 and that's very simply zero. Zero is a very legitimate answer. We're not dealing with a zero divided by zero or one divided by zero or anything like that. Simple direct substitution, simple solution.