Why is lim f(x)×g(x) indeterminate when f(x)→∞ and g(x)→0?

[negation]

Homework Statement

Suppose the lim x-> a f(x) = + infinity and lim x->a g(x) = 0

then why would lim x→a (f(x)×g(x)) be insufficient to tell us anything about the product of 2 limit?

 

[skrat]

Because the product of $\infty \times 0$ is not defined.

Write $g(x)f(x)=\frac{g(x)}{\frac{1}{f(x)}}$ and now calculate the limit.

 

[Ray Vickson]

Homework Statement

Suppose the lim x-> a f(x) = + infinity and lim x->a g(x) = 0

then why would lim x→a (f(x)×g(x)) be insufficient to tell us anything about the product of 2 limit?

Show your work. Where are you stuck?

 

[LCKurtz]

Homework Statement

Suppose the lim x-> a f(x) = + infinity and lim x->a g(x) = 0

then why would lim x→a (f(x)×g(x)) be insufficient to tell us anything about the product of 2 limit?

Look at these two examples taking $a=0$ so $x\rightarrow 0^+$:

$f(x) = \frac 1 x \rightarrow \infty,~g(x) = x^2\rightarrow 0$. Here $f(x)g(x) = x\rightarrow 0$

Now take $f(x) = \frac 1 {x^2}\rightarrow \infty,~g(x) = x\rightarrow 0$. Here $f(x)g(x)= \frac 1 x\rightarrow \infty$.

So you having $f\to\infty,~g\to 0$ isn't sufficient to tell us anything about $fg$.

 

[HallsofIvy]

And to complete what LCKurtz said, for a any non-zero number,
If $f(x)= \frac{a}{x}$ and $g(x)= x$, then $\lim_{x\to 0} f(x)g(x)= a$

So that, in fact, there are examples giving every possible result!

 

[negation]

Look at these two examples taking $a=0$ so $x\rightarrow 0^+$:

$f(x) = \frac 1 x \rightarrow \infty,~g(x) = x^2\rightarrow 0$. Here $f(x)g(x) = x\rightarrow 0$

Now take $f(x) = \frac 1 {x^2}\rightarrow \infty,~g(x) = x\rightarrow 0$. Here $f(x)g(x)= \frac 1 x\rightarrow \infty$.

So you having $f\to\infty,~g\to 0$ isn't sufficient to tell us anything about $fg$.

This is tough to grasp or maybe I'm missing some intermediate steps.

By the basic limit law of multiplication:

lim x->a f(x) and lim x->a g(x)
then lim x->a f(x) . lim x->a g(x) = lim x->a f(x) .g(x)

so f(x) -> infinity and g(x) -> f(x).g(x) = 0

 

[Mark44]

This is tough to grasp or maybe I'm missing some intermediate steps.

By the basic limit law of multiplication:

lim x->a f(x) and lim x->a g(x)

Equal what?

From post #1, $lim_{x \to a} f(x) = \infty$ and $lim_{x \to a} g(x) = 0$.

then lim x->a f(x) . lim x->a g(x) = lim x->a f(x) .g(x)

The property you are citing, about the multiplication of limits, requires that both limits exist. That means that each limit has to be a finite number. So your first limit does not exist.

so f(x) -> infinity and g(x) -> f(x).g(x) = 0

No.
There are already several examples in this thread that show that this is not a valid conclusion.

 

[skiller]

And to complete what LCKurtz said, for a any non-zero number,
If $f(x)= \frac{a}{x}$ and $g(x)= x$, then $\lim_{x\to 0} f(x)g(x)= a$

So that, in fact, there are examples giving every possible result!

Why complicate things by mixing up a and zero?

 

[HallsofIvy]

Why complicate things by mixing up a and zero?

I don't know what you mean by this. Mixing up "a" and what "zero"?

 

[Mark44]

Why complicate things by mixing up a and zero?

HallsOfIvy is making the point that an [∞ * 0] indeterminate form can turn out to be any number.

 

[skiller]

I don't know what you mean by this. Mixing up "a" and what "zero"?

I believe there was only one zero in your post - that zero was the one you apparently don't see as the problem.

The OP's question was to do with the limit as $x$ tended to $a$. Your comment was to do with $x$ tending to $0$ with an $a$ in the expression - a very different $a$!

It may be considered to be a small pedantic point, but I think it's very important and to my mind you added unnecessary confusion. Perhaps your mind was swayed by LCKurtz's example of $a$ being $0$.

HallsOfIvy is making the point that an [∞ * 0] indeterminate form can turn out to be any number.

Thanks, I was aware of that.