2. Limits’ properties and related theorems
We can see that it’s really hard to find the limit of a complex function by using formal definition directly. Hence, here is some limits’ properties and theorems that can be useful in finding limit. Firstly, let’s rewrite definition of limits:
2.1. Limit laws for functions
Proof: (a) It suffices to prove , i.e. according to definition 11 for any there is so for all . We need to use given condition that and , i.e. for any there exists so that for all and for all . Note that
Hence, we can pick so that . This will follow for all . Thus, (a) is proven according to definition 11.
(b) It suffices to prove . The case where is obvious since we can choose which follows equals to either or which are both less than for all .
WLOG, assume that . It suffices to prove that for every there is a so for all . In order to do this, we need to bring back to familiar where we already knew the limits. Hence, we can pick so that for in some interval to get in that interval. Indeed, note that according to definition 11 then for all and for all . Therefore, we can choose so or , which is possible since . Hence, for all then .
Therefore, for all , we can pick and so . With this, for all so then so
According to definition 11, for .
(c) If then it’s obvious. We only need to consider . It suffices to prove , i.e. for any there exists so for all . Pick then since so for there exists so or for all . We are done.
(d) It suffices to prove , i.e. for any there exists so for all . We need to use condition and to imply this, i.e. for all and for all . We need to apear from and , so we can multiply the two to get
Now we have , the next thing is to manipulate the so that can be easily choosen to make arbitrary small. Indeed, we have
A remark that in order to use above limit laws, we need a condition that and to be defined or else the limit laws will not work. Evidently, we can see that all of the proofs represented above use the condition that and are defined to prove the limit laws. Intuitively, if or are not defined then obviously we can’t calculate from .
Theorem 12 is missing one more limit law, which is limit of division:
What condition of can we use (3)? Simiarly to above argument, of course and must be defined. But is that condition good enough to use (3)? Notice that in (3) we have division by so one more condition need to be added is in order to calculate RHS of (3). Hence, we obain the following theorem:
With we guarantee that RHS of (3) can be calculated. Can guarantee that limit in LHS of (3) satisfy prerequisit condition for a limit to be defined, i.e. the function must be defined on some interval containing except possibly at (see definition 11)? Yes, it can. We can show that is equivalent to for all . Indeed, if then for any there exists so for all . With this, if there exists so then , a contradiction. Thus, for all then . The reverse is also true. Hence,
for all with and contains .
Combining this with condition that is defined on , we find is defined on . Thus, the prerequisit condition is satisfied. Hence, we now understand why is added to the theorem’s condition. We also know that we can replace condition with for with contains .
Now, let’s try and prove theorem 13.
Proof: If we look at the proof of theorem 12 (d) and try to use the similar idea, we see that it would be harder to manipulate and to bring back . Instead, we can prove and then use theorem 12 (d) to say
We can see that proving is a much easier job. Indeed, note that in above argument, we find that is defined on containing . Since , we consider an so for all . We have and note that can be negative number. We notice that for any from this. Hence, for all then
Let’s just write then . Thus, we can find from any . Therefore, for any then for all . This follows . From this and theorem 12, we obtain the desired result.
2.2. Left, right limit theorem
The proof for this theorem can be obtained directly from definition 5 of one-side limits.
2.3. Squeeze theorem
Proof: This is straight off from definition. For any there exists so for all and for all . Hence, we find for any then
for all . This follows .
Let’s go through a classic example using Squeeze theorem:
Geometric solution combining eith Squeeze theorem for this problem can be found on this link. In there, they provide two different geometric approaches to establish the bound
One solution use areas and the other use comparison between lengths in unit circle. The second solution points out that the limit explicitly depends on being measured in radians.
2.4. Squeeze theorem for limits involving or
When the limit involves with or , we often think of L’Hopital’s Rule. However, we can also use Squeeze theorem to prove some limits of this type. In particular, we need to know two inequalities