# Limit of 1/x, as x approaches inf, question.

1. Dec 23, 2008

### ellis818

Multiplication Question

If we know that $$\frac{a}{x}$$ = c, we than say a=x*c

We also know that $$\lim_{x\rightarrow \infty}$$ $$\frac{a}{x}$$ = 0

But what if we said: $$\frac{a}{x}$$ = c, then $$\frac{a*x}{x}$$ = x*c

So, if we take a limit: $$\lim_{x\rightarrow \infty}$$ $$\frac{a*x}{x}$$ = a

And also take a limit of: $$\lim_{x\rightarrow \infty}$$ x*c = $$\infty$$

We get a problem.

Last edited: Dec 23, 2008
2. Dec 23, 2008

### rock.freak667

Could you make your question a bit more specific?

3. Dec 23, 2008

### jgens

Is the question supposed to be: lim (x -> infinity) 1/x?

4. Dec 23, 2008

### ellis818

Sorry, accidently hit the Post button without being finished.

5. Dec 23, 2008

### jgens

I think that the resolution of the conflict arises in the fact that lim (x -> infinity) cx = infinity when c is not equal to zero. I think in this particular case lim (x -> infinity) cx is an indeterminate form in the form of 0*infinity. I could be entirely wrong though.

6. Dec 23, 2008

The problem seems to be that c depends on x. Remember, c = x / a, so
$$\lim_{x \to \infty} cx = \lim_{x \to \infty} \frac{x}{a} x = \lim_{x \to \infty} a = a.$$

7. Dec 23, 2008

### ellis818

Could you please explain this a bit further?

8. Dec 23, 2008

Whoops, I made a typo. It should be c = a / x and

$$\lim_{x \to \infty} cx = \lim_{x \to \infty} \frac{a}{x} x = \lim_{x \to \infty} a = a.$$

9. Dec 23, 2008

### ellis818

Actually what I think you meant was, $$\lim_{x \to \infty} \frac{a}{x} x$$

10. Dec 23, 2008

### ellis818

adriank I understand what you did, but the question I have is that both sides are not valid. I wasn't looking for a substatution of c being a/x.

11. Dec 23, 2008

I'm using what you used for c in your original post. The point is, in your post, c is not a constant.

12. Dec 23, 2008

### ellis818

That's what I was missing. Thanks adriank.

13. Dec 23, 2008

### ellis818

This leads me to another question, somewhat relevant to the original post.

If $$\lim_{x \to \infty} x (\lim_{x \to \infty}\frac{a}{x})$$ = a

Does this sort of mean that $$\infty$$*0 = a ?

a is any real number except 0

14. Dec 23, 2008

### sutupidmath

i would say that $$\lim_{x \to \infty} x (\lim_{x \to \infty}\frac{a}{x})$$ is an intermediate form, that is $$\infty*0$$ so as it is right now, we cannot evaluate the limit, but if we transform it into

$$\lim_{x \to \infty} x (\lim_{x \to \infty}\frac{a}{x})=\lim_{x\rightarrow \infty}x\frac{a}{x}=a$$

Remember that infinity is not a number, so the laws that hold for real numbers, not necessarly will hold when infinity is in play.

15. Dec 23, 2008

### ellis818

Well basically my argument was the if we take the seperate parts of the two being:

$$\lim_{x \to \infty} x = \infty$$

and

$$\lim_{x \to \infty}\frac{a}{x} = 0$$

multiply them together and get

$$\lim_{x \to \infty}\frac{a}{x}x = a$$

16. Dec 23, 2008

Your equation is ambiguous; there are two possible interpretations. First, and I'm guessing this is what you meant:
$$\left(\lim_{x \to \infty} x\right) \left(\lim_{x \to \infty}\frac{a}{x}\right) = a.$$
The first limit doesn't exist, so that doesn't work. (Infinite limits are said to not exist; infinity is not a number. You can't multiply by it.)

Remember that in general,
$$\lim_{x \to \infty} f(x) g(x) = \left( \lim_{x \to \infty} f(x) \right) \left( \lim_{x \to \infty} g(x) \right)$$
is true only if each of the limits exists.

Second:
$$\lim_{x \to \infty} \left(x \lim_{x \to \infty}\frac{a}{x}\right) = a.$$
That just doesn't make sense. It isn't meaningful. You can't use the same variable in a limit that's inside another limit; it's just wrong to write that.

Perhaps you meant this:
$$\lim_{y \to \infty} \left(y \lim_{x \to \infty}\frac{a}{x}\right).$$
In that case, evaluate the limits one at a time:
$$\lim_{y \to \infty} \left(y \lim_{x \to \infty}\frac{a}{x}\right) = \lim_{y \to \infty} \left(y \cdot 0\right) = \lim_{y \to \infty} 0 = 0.$$

17. Dec 24, 2008

### ellis818

Your right. I should have noticed that. I guess I was lazy to check that idea in the first place.