Limit existence rules

The basic limit laws are stated below: \begin{align*} \lim [f+g]&=\lim f + \lim g \\ \lim[f.g]&=\lim f.\lim g \\ \lim[\frac{f}{g}]&=\frac{\lim f}{\lim g} \end{align*}

I'm trying to state some basic rules about the existence of such limits. For example it is sometimes mistakenly thought that if $\lim f, \lim g$ both do not exist then $\lim [f+g]$ cannot exist. My question is, are the following statements true:

1. If any two limits exist then third exists except in cases where we must divide by $0$

2. If two limits do not exist, it may be the case the third exist.

3. Generally, if one exists and the other does not exist, the third does not exist. Except:

a)If $\lim f$ is equal to $0$ and $\lim g$ does not exist, then $\lim[f.g]$ then may exist;

b)If lim g is infinity and lim f exists, then lim [f/g] exists.

And if not true, why? As you can see I'm trying to generate a schemata about limit existence so any further help would be appreciated.

• Jun 15 '18 at 14:30
• @helios321 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Aug 4 '18 at 22:10

For point 1 we have that if the two limits exist also the third exists with some exception. For the sum the exception is when both $f$ ang $g$ tend to $\infty$ with opposite sign and for the ratio when $g$ tends to zero.

For the second point just consider the limit $x\to \infty$ for

• $f(x)=\sin x$
• $g(x)=-\sin x$

the limit of the sum exists and

• $f(x)=\sin x$
• $g(x)=\frac1{\sin x}$

and the limit of the product exists.

For the third point you have already find some good examples.

• For your first point what is the exception? Your example is where $f$ and $g$ do not exist which is not my hypothesis in 1. Jun 15 '18 at 5:03
• That’s a matter of definition for “the limit exists”, often we consider that the limit exist also when it is $\infty$ and we refer to a not existing limit for the cases as $\sin x$ for $x\to \infty$. If you are using a different definition of course the only exception is for $g\to 0$ for the ratio.
– user
Jun 15 '18 at 5:23