# A question about a limit

Note: all the following limits are when x approaches some certain number

The question is: if lim(f(x)g(x)) doesn't exist, and lim(f(x)) = 1, and lim(g(x)) doesn't exist
then are we allowed to say that lim(f(x)g(x)) = lim(f(x)) * lim (g(x)) = 1 * lim(g(x)) = doesn't exist

or isn't that allowed in general

Thanks alot

chwala
Gold Member
if lim (f(x)) is continous and lim (g(X)) is not then it follows that:
lim(f(x)*g(x)) is not continous...for the simple reason that f(x) is bounded and g(x) is not bounded...for e.g if f(x)= 1/n then its and n→∞ then the lim→0 and conversely for g(x)=n when n→∞ lim→∞

What you're saying is allowable, but I want to tell you that it's useless and makes absolutely no sense to repeat yourself 3 times. I'm not a pro on limits but you should understand what you're doing...

The simple reason is because all your doing is stating what is equal three times

lim(f(x)g(x) = lim(f(x)) *limg(x)) All you did was show us another way to write how to multiply a function

then you say that 1 * lim(g(x)) =doesn't exist

Of course not

if f(x) = 1 all you did was plug in a 1 instead of F(x)

You mine as well just said

lim(f(x)g(x)) = lim(f(x)) * lim (g(x)) = 1 * lim(g(x)) = limf(x) * limg(x) = lim(f(x)g(x))
^ If you would've just worked your way back around you would've made it.

Essentially all you did was substitution, but yeah not really a limit question more of a algebra question

chwala
Gold Member
What you're saying is allowable, but I want to tell you that it's useless and makes absolutely no sense to repeat yourself 3 times. I'm not a pro on limits but you should understand what you're doing...

The simple reason is because all your doing is stating what is equal three times

lim(f(x)g(x) = lim(f(x)) *limg(x)) All you did was show us another way to write how to multiply a function

then you say that 1 * lim(g(x)) =doesn't exist

Of course not

if f(x) = 1 all you did was plug in a 1 instead of F(x)

You mine as well just said

lim(f(x)g(x)) = lim(f(x)) * lim (g(x)) = 1 * lim(g(x)) = limf(x) * limg(x) = lim(f(x)g(x))
^ If you would've just worked your way back around you would've made it.

Essentially all you did was substitution, but yeah not really a limit question more of a algebra question
spped,
was i right in my response...thanks and goodnight

chwala

Thanks guys

but I want to tell you that it's useless and makes absolutely no sense to repeat yourself 3 times

If you just knew why I asked this questions, you wouldn't have said that
I have a limit in my exam which is lim( ((x^2 - 2x +1) ^ 0.5) / (x(x-1))) as x approaches one
in the end the limit doesn't exist, what I did is separate this to two limits, one of which is lim 1/x
and continued my solution properly
without this separation I would have taken a full mark in my exam, but my teacher said it's not allowable because one of these limits doesn't exist,
so he took from me a mark and had given me a big bang :D

I'm insisting to convince him so help me ;)

thanks

The problem is, you're assuming what you want to prove.

Splitting up only works if you know that the limit already doesn't exist, in which case it defeats the purpose of the question. For example,

$$1 = \lim_{x \rightarrow 0} 1 = \lim_{x \rightarrow 0} \frac{x}{x} = \lim_{x \rightarrow 0} x \cdot \frac{1}{x}$$

If you split it it, you get one of the factors that doesn't exist, but the limit itself exists. You have to be more careful when making such algebra tricks, which is why your prof took off points. Too careless. : )

if lim (f(x)) is continous and lim (g(X)) is not then it follows that:
lim(f(x)*g(x)) is not continous...for the simple reason that f(x) is bounded and g(x) is not bounded...for e.g if f(x)= 1/n then its and n→∞ then the lim→0 and conversely for g(x)=n when n→∞ lim→∞

Hey chwala.. probably u have to be more careful... u said f(x) has limit because it is bounded... now consider $$(-1)^{n}$$ .. But im sure what u've state about g(x) exisiting no limits before it is not bounded ..

chwala
Gold Member
Hey chwala.. probably u have to be more careful... u said f(x) has limit because it is bounded... now consider $$(-1)^{n}$$ .. But im sure what u've state about g(x) exisiting no limits before it is not bounded ..

thanks you are right....(-1)^n is bounded by -1 and 1 but has no limit i.e if n→∞ the limit does not exist...i withdraw the statement on the boundedness of f(x).

The problem is, you're assuming what you want to prove.

Splitting up only works if you know that the limit already doesn't exist, in which case it defeats the purpose of the question. For example,

$$1 = \lim_{x \rightarrow 0} 1 = \lim_{x \rightarrow 0} \frac{x}{x} = \lim_{x \rightarrow 0} x \cdot \frac{1}{x}$$

If you split it it, you get one of the factors that doesn't exist, but the limit itself exists. You have to be more careful when making such algebra tricks, which is why your prof took off points. Too careless. : )

mustn't there be a difference between a non-existing limit and an undefined limit?