Assume that  and  exist and that c is any constant.  Then,

This is a very simple proof.  To make the notation a little clearer let’s define the function  then what we’re being asked to prove is that .  So let’s do that.

Let  and we need to show that we can find a  so that

The left inequality is trivially satisfied for any x however because we defined .  So simply choose  to be any number you want (you generally can’t do this with these proofs).  Then,

There are several ways to prove this part.  If you accept 3 And 7 then all you need to do is let  and then this is a direct result of 3 and 7.  However, we’d like to do a more rigorous mathematical proof.  So here is that proof.

First, note that if  then  and so,

The limit evaluation is a special case of 7 (with  ) which we just proved  Therefore we know 1 is true for  and so we can assume that  for the remainder of this proof.

Let  then because  by the definition of the limit there is a  such that,

Now choose  and we need to show that

and we’ll be done.  So, assume that  and then,

We’ll be doing this proof in two parts.  First let’s prove .

Let  then because  and  there is a  and a  such that,

Now choose .  Then we need to show that

Assume that .  We then have,

In the third step we used the fact that, by our choice of , we also have  and  and so we can use the initial statements in our proof.

Next, we need to prove .  We could do a similar proof as we did above for the sum of two functions.  However, we might as well take advantage of the fact that we’ve proven this for a sum and that we’ve also proven 1.

This one is a little tricky.  First, let’s note that because  and  we can use 2 and 7 to prove the following two limits.

Now, let .  Then there is a  and a  such that,

Choose .  If  we then get,

So, we’ve managed to prove that,

This apparently has nothing to do with what we actually want to prove, but as you’ll see in a bit it is needed.

Before launching into the actual proof of 3 let’s do a little Algebra.  First, expand the following product.

Rearranging this gives the following way to write the product of the two functions.

With this we can now proceed with the proof of 3.

Fairly simple proof really, once you see all the steps that you have to take before you even start.  The second step made multiple uses of property 2.  In the third step we used the limit we initially proved.  In the fourth step we used properties 1 and 7.   Finally, we just did some simplification.

This one is also a little tricky.  First, we’ll start of by proving,

Let  .  We’ll not need this right away, but these proofs always start off with this statement.  Now, because  there is a  such that,

Now, assuming that  we have,

Rearranging this gives,

Now, there is also a  such that,

Choose .  If  we have,

Now that we’ve proven  the more general fact is easy.

As noted we’re only going to prove 5 for integer exponents.  This will also involve proof by induction so if you aren’t familiar with induction proofs you can skip this proof.

So, we’re going to prove,

For  we have nothing more than a special case of property 3.

So, 5 is proven for .  Now assume that 5 is true for , or .  Then, again using property 3 we have,

As pointed out in the Limit Properties section this is nothing more than a special case of the full version of 5 and the proof is given there and so is the proof is not give here.

This is a simple proof.  If we define  to make the notation a little easier, we’re being asked to prove that .

Let  and let .  Then, if  we have,

 So, we’ve proved that .

This is just a special case of property 5 with  and so we won’t prove it here.

Given the functions  and  suppose we have,

for some real numbers c and L.  Then,

1.  

2. If  then  

3. If  then  

4.  

We will prove  here.  The proof of  is nearly identical and is left to the reader.

Let  then because we know  there exists a  such that if  we have,

Also, because we know  there exists a  such that if  we have,

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

 So, we’ve proved that .

Let  then because we know  there exists a  such that if  we have,

Also, because we know  there exists a  such that if  we have,

Note that because we know that  choosing  in the first inequality above is a valid choice because it will also be positive as required by the definition of the limit.

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

 So, we’ve proved that .

Let  then because we know  there exists a  such that if  we have,

Note that because  in the case we will have  here.  

Next, because we know  there exists a  such that if  we have,

Again, because we know that  we will have .  Also, for reasons that will shortly be apparent, multiply the final inequality by a minus sign to get,

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

This may seem to not be what we needed however multiplying this by a minus sign gives,

 and because we originally chose  we have now proven that .

We’ll need to do this in three cases.  Let’s start with the easiest case.

Case 1 :  

Let  then because we know  there exists a  such that if  we have,

Next, because we know  there exists a  such that if  we have,

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

In the second step we could remove the absolute value bars from  because we know it is positive.

So, we proved that  if .

Case 2 :  

Let  then because we know  there exists a  such that if  we have,

Next, because we know  there exists a  such that if  we have,

Also, because we are assuming that  it is safe to assume that for  we have .

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

In the second step we could remove the absolute value bars because we know or can safely assume (as noted above) that both functions were positive.

So, we proved that  if .

Case 3 : .

Let  then because we know  there exists a  such that if  we have,

Next, because we know  there exists a  such that if  we have,

Next, multiply this be a negative sign to get,

Also, because we are assuming that  it is safe to assume that for  we have .

Now, let  and so if  we know from the above statements that we will have both,

This gives us,

In the second step we could remove the absolute value bars by adding in the negative because we know that  and can safely assume that  (as noted above).

So, we proved that  if .

1. If r is a positive rational number and c is any real number then,

2. If r is a positive rational number, c is any real number and x^r  is defined for  then,

This is actually a fairly simple proof but we’ll need to do three separate cases. 

Case 1 : Assume that . Next, let  and define .  Note that because c and  are both positive we know that this root will exist.  Now, assume that we have .  Give this assumption we have,

So, provided  we’ve proven that .

Case 2 : Assume that .  Here all we need to do is the following,

Case 3 : Finally, assume that .  In this case we can then write  where .  Then using Case 1 and the fact that we can factor constants out of a limit we get,

This is very similar to the proof of 1 so we’ll just do the first case (as it’s the hardest) and leave the other two cases up to you to prove.

Case 1 : Assume that . Next, let  and define .  Note that because c and  are both positive we know that this root will exist.  Now, assume that we have .  Note that this assumption also tells us that x will be negative. Give this assumption we have,

So, provided  we’ve proven that .  Note that the main difference here is that we need to take the absolute value first to deal with the minus sign.  Because both sides are negative we know that when we take the absolute value of both sides the direction of the inequality will have to switch as well.

Case 2, Case 3 : As noted above these are identical to the proof of the corresponding cases in the first proof and so are omitted here.

If  is a polynomial of degree n (i.e.  )  then,

We’re going to prove this in an identical fashion to the problems that we worked in this section involving polynomials.  We’ll first factor out  from the polynomial and then make a giant use of Fact 1 (which we just proved above) and the basic properties of limits.

Now, clearly the limit of the second term is one and the limit of the first term will be either  or  depending upon the sign of .  Therefore by the Facts from the Infinite Limits section we can see that the limit of the whole polynomial will be the same as the limit of the first term or,

The proof of this part is literally identical to the proof of the first part, with the exception that all  ’s are changed to , and so is omitted here.

If  is continuous at  and  then,

Let  then we need to show that there is a  such that,

Let’s start with the fact that  is continuous at .  Recall that this means that  and so there must be a  so that,

Now, let’s recall that .  This means that there must be a  so that,

But all this means that we’re done. 

Let’s summarize up.  First assume that .  This then tells us that,

But, we also know that if  then we must also have .  What this is telling us is that if a number is within a distance of  of b then we can plug that number into  and we’ll be within a distance of  of .

So,   is telling us that  is within a distance of  of b and so if we plug it into  we’ll get,

and this is exactly what we wanted to show.