# Thread: Limit property doesn't seem to ring true

1. ## Limit property doesn't seem to ring true

Taking the limit as x approaches infinity of: $\frac{x^2}{x-1}-\frac{x^2}{x+1}$ can be rewritten as:

The limit as x approaches infinity of: $\frac{x^2}{x-1}$ MINUS the limit as x approaches infinity of: $\frac{x^2}{x+1}$

But this isn't true.

If I re-arrange algebraically in the first example, I will get x=2.

However, if I solve the rewritten problem, I get infinity minus infinity or undefined...??

2. ## Re: Limit property doesn't seem to ring true

Originally Posted by Paze
Taking the limit as x approaches infinity of: $\frac{x^2}{x-1}-\frac{x^2}{x+1}$ can be rewritten as:
The limit as x approaches infinity of: $\frac{x^2}{x-1}$ MINUS the limit as x approaches infinity of: $\frac{x^2}{x+1}$
$\frac{x^2}{x-1}-\frac{x^2}{x+1}=\frac{(x^3+x^2)-(x^3-x^2)}{x^2-1}=\frac{2x^2}{x^2-1}$

3. ## Re: Limit property doesn't seem to ring true

The theorem that says that $\lim_{x\to\infty}(f(x)-g(x))= \lim_{x\to\infty}f(x)-\lim_{x\to\infty}g(x)$ has a proviso saying that the limits in the right-hand side exist. In this situation, the theorem does not apply.

4. ## Re: Limit property doesn't seem to ring true

Originally Posted by emakarov
The theorem that says that $\lim_{x\to\infty}(f(x)-g(x))= \lim_{x\to\infty}f(x)-\lim_{x\to\infty}g(x)$ has a proviso saying that the limits in the right-hand side exist. In this situation, the theorem does not apply.
I'm sorry, but I don't understand. Why doesn't my right hand side limit 'exist'?

5. ## Re: Limit property doesn't seem to ring true

Originally Posted by Paze
I'm sorry, but I don't understand. Why doesn't my right hand side limit 'exist'?
I don't know what your textbook says about the validity of that theorem. The previous post informed you under what condition the theorem is valid. If you still have any doubt about it, perhaps you would need to go back to the proof of the throrem step by step.

6. ## Re: Limit property doesn't seem to ring true

Originally Posted by votan
I don't know what your textbook says about the validity of that theorem. The previous post informed you under what condition the theorem is valid. If you still have any doubt about it, perhaps you would need to go back to the proof of the throrem step by step.
I don't understand what "the right hand side needs to exist" means. My text-book is in another language

7. ## Re: Limit property doesn't seem to ring true

Originally Posted by Paze
I don't understand what "the right hand side needs to exist" means. My text-book is in another language
I sense your confusion stems from language barrier. if you say lim(f(x)) as x-->n, exist means at the limit f(n) is finite, or can be resolve without anbguity. if your expression leads to (inf) + (inf), that limit cannot be resolve. You will have to rearrange your expression to an equivalent form. if now the limit is finite, then that limit exists, but if it is still infinte, than you say the limit does not exist. Hopefully this helps.

8. ## Re: Limit property doesn't seem to ring true

There is a theorem that says "IF $\lim_{x\to a} f(x)$ exists and $\lim_{x\to a} g(x)$ exists then $\lim_{x\to a} (f(x)+ g(x)= [\lim_{x\to a} f(x)]+ [\lim_{x\to a} g(x)]$.

You may be thinking that " $\lim_{x\to a} f(x)= \infty$" means that the limit exists and is infinity. That is not the case. Saying that " $\lim_{x\to a} f(x)= \infty$" means "the limit does NOT exist" (in a particular way).

9. ## Re: Limit property doesn't seem to ring true

I see! Thanks guys!

10. ## Re: Limit property doesn't seem to ring true

Originally Posted by HallsofIvy
There is a theorem that says "IF $\lim_{x\to a} f(x)$ exists and $\lim_{x\to a} g(x)$ exists then $\lim_{x\to a} (f(x)+ g(x)= [\lim_{x\to a} f(x)]+ [\lim_{x\to a} g(x)]$.

You may be thinking that " $\lim_{x\to a} f(x)= \infty$" means that the limit exists and is infinity. That is not the case. Saying that " $\lim_{x\to a} f(x)= \infty$" means "the limit does NOT exist" (in a particular way).
I re-read my post. It doesn't mean to me that the limit exists and is infinity.