Limit of product of functions when one of them has no limit

[sari]

Homework Statement

If it is known that f has no limit (finite or infinite) at x, does the limit if f*g at x exist? (g being any continuous function). if not - how do you prove it?

 

[SteamKing]

If f has no limit at x, the limit of f*g at x does not exist. This can be proved by the algebraic limit theorem.

See http://en.wikipedia.org/wiki/Limit_of_a_function#Chain_rule

 

[SammyS]

If f has no limit at x, the limit of f*g at x does not exist. This can be proved by the algebraic limit theorem.

See http://en.wikipedia.org/wiki/Limit_of_a_function#Chain_rule

What if g has a limit of zero at x ? Then the limit of f*g at x may or may not be zero, depending upon the details of the functions f & g.

 

[dynamicsolo]

Homework Statement

If it is known that f has no limit (finite or infinite) at x, does the limit if f*g at x exist? (g being any continuous function). if not - how do you prove it?

If I read this correctly, you are asking about the situation where \lim_{x \rightarrow a} f(x) literally does not exist. To expand a bit on what SammyS said, there are product functions for which \lim_{x \rightarrow a} f(x) has no value, yet \lim_{x \rightarrow a} f(x) g(x) has a meaningful value. (Admittedly, many of these functions are a bit contrived.)

The classic example is \lim_{x \rightarrow 0} \sin( \frac{1}{x} ) , which does not exist, but \lim_{x \rightarrow 0} x \sin( \frac{1}{x} ) = 0 (and, indeed, \lim_{x \rightarrow 0} x^{p} \sin( \frac{1}{x} ) = 0 , for p > 0 ) . Functions like this are ones that the "Squeeze Theorem" (or "Sandwich Theorem") were invented to find limits for, since there is no computational method of assessing the limit of a product when one of the limits is undefined.Side question: why is TeX refusing to display the expressions in my last paragraph properly? (FIXED)

 

[sari]

Right - that's exactly the situation in my problem. lim g = 0, lim f doesn't exist.

Specifically, I'm trying to determine whether x^2* (1/sin(1/x)) has a limit at 0.

1/sin(1/x) oscillates around between +-infinity in the neighborhood of 0, so it clearly has no limit. However the limit of x^2 at 0 is 0.

 

[sari]

dynamicsolo - thanks for the help! Didn't notice your answer before I finished my post!

 

[sari]

Oh, wait - I'm still stuck! Obviously, lim x^p*sin(1/x) = 0 at 0, because sin(1/x) is bounded between +-1.

But what do you do when you have a case like x^p*1/sin(1/x) ? Squeeze theorem doesn't seem to work here.

 

[dynamicsolo]

Right - that's exactly the situation in my problem. lim g = 0, lim f doesn't exist.

Specifically, I'm trying to determine whether x^2* (1/sin(1/x)) has a limit at 0.

1/sin(1/x) oscillates around between +-infinity in the neighborhood of 0, so it clearly has no limit. However the limit of x^2 at 0 is 0.

Hmm, never saw that function before. Of course, the "Squeeze Theorem" is fine for
x^{2} \sin ( \frac{1}{x} ) , but the function \frac {x^{2}}{ \sin ( \frac{1}{x} ) } has an infinite number of vertical asymptotes of increasing density as x approaches zero. I suspect the answer here is that the limit approaching zero does not exist because there is no finite number which is approached "arbitrarily closely" by the function and the function "runs off" to both "positive" and "negative infinity" with increasing density, so we don't even have a single "infinite limit". So there is no meaningful way to talk about what this function does as x approaches zero.

 

[sari]

I guess you could try to find two different series approaching 0 for which the function approaches different values. I'm going to try that approach...

 

[SammyS]

There was a space in each of your [/tex ] tags.

 

[dynamicsolo]

There was a space in each of your [/tex ] tags.

Thank you! I've fixed that and will be more careful about what I've typed. (Man, these computers are sooo fussy...)

 

[snipez90]

Why don't you just let f(x) = 1/x.

 

[dynamicsolo]

Why don't you just let f(x) = 1/x.

It's true that 1/x has no "two-sided limit" as x approaches zero, so the limit \lim_{x \rightarrow 0} \frac{1}{x} doesn't exist. I think the function sari was being asked about, \lim_{x \rightarrow 0} \frac{1}{\sin (\frac{1}{x}) } was chosen to illustrate the issue more emphatically...