# Thread: A question on limits

1. ## A question on limits

Suppose limf(x)=limg(x) as x->a.
Now:
Are we always permitted to subsituate f(x) with g(x) whenever we calculate a limit(consiting f or g) as x->a?
Why or why not?(wheter yes or not, Prove it - not by exmple)
Regards.

2. Well, I would say no, because in some problems, it matters a great deal how the functions behave near the limit, not just what the limiting value is. So here's a proposed counterexample:

Let $\displaystyle f(x)=x$ and
$\displaystyle g(x)=x^{2}.$

Note that

$\displaystyle \displaystyle{\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=0.}$

However,

$\displaystyle \displaystyle{1=\lim_{x\to 0}\frac{x}{f(x)}\not=\lim_{x\to 0}\frac{x}{g(x)},}$ which does not exist.

This may or may not be a counterexample. It depends on what you mean by the phrase "consisting [of] f or g". I've got an example that has f and g in it, and I've shown that you can't just swap them out and expect to get the same thing.

3. Thanks.

But I know lots of another counterexamples for it and that's why i asked a clear reason or proof!

Of course there exists a rule that lets us to do so by some specific conditions.
But unfortunately i couldn't find it on english books and articles about basic and advanced calculus!Only found it (without its proof) and its application on a farsi book about limits.
After our discussion i'll show you that if you wish!

But now the important question is why can't we do what i mentioned at first; and it requires a clear reason...

4. If I state a hypothesis that is either true or false, and I give a counterexample, that is a proof of falsity. For example, if I claim that every integer is divisible by 2, and then produce 3 as a counterexample, I have just proven (no need for a clearer proof!) that it is not the case that every integer is divisible by 2.

But unfortunately I couldn't find it on english books and articles about basic and advanced calculus!
Herein lies part of the problem, I think. I don't think you've defined clearly enough the problem you're trying to solve. This is what I meant when I said that the phrase "consisting [of] f or g" is vague. I've shown clearly, without need for further proof, that the equality of two limits does not imply the equality of other limits involving the functions used in those two limits. But that may or may not be what you meant by "consisting of f or g". Until you clearly define that, I'm not sure I can be of much help to you.

5. Sorry you are right; That's a proof! But its reason is needed....
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To know the meaning of what i say note the following examle:
We want to evaluate lim [f(x)+A(x)/2L(x)^5] as x->a, and we have lim f(x)=lim g(x) as x->a.
Therefore by limits laws we have: lim [f(x)+A(x)/2L(x)^5]=[lim f(x)+lim A(x)]/[lim 2L(x)^5] as x->a.
Now because of the blue line we could replace the limit of f with the limit of g and change the limit as [lim g(x)+lim A(x)]/[lim 2L(x)^5] as x->a and then solve it(if possible!)
Infact i want to know the reason that we can not do always like that.
We know that they are equal and that two equal things can be replaced in an expression.
So why they can't be so?! What is the reason of this inconsistency?
Thanks

6. I suppose the reason for the inconsistency is because of how FAST the functions approach their limits. You know how 2 functions may intersect at a few points but then not be equal at other points. It is the same with limits. Two functions may have the same limit at Xo but then fail to have the same limit at another x.

As for when you can do the substitution that you showed, I think that you can make the procedure more rigorous. For now, an example should suffice.

If $\displaystyle \displaystyle \underset{x \to x_0}{\lim} \frac{f(x)+A(x)}{2L(x)^5}= \frac{\underset{x \to x_0}{\lim}f(x)+\underset{x \to x_0}{\lim}A(x)}{\underset{x \to x_0}{\lim} 2L(x)^5}$

then $\displaystyle \displaystyle \underset{x \to x_0}{\lim} \frac{f(x)+A(x)}{2L(x)^5}= \frac{\underset{x \to x_0}{\lim}g(x)+\underset{x \to x_0}{\lim}A(x)}{\underset{x \to x_0}{\lim} 2L(x)^5}$

A more general version would be

Let $\displaystyle h(x,c)$ be a function. If $\displaystyle \underset{x \to x_o}{\lim}f(x) = \underset{x \to x_o}{\lim}g(x)$ and $\displaystyle \displaystyle \underset{x \to x_o}{\lim}h(x,f(x)) = h(x_o, \underset{x \to x_o}{\lim}f(x))$

then $\displaystyle \displaystyle \underset{x \to x_o}{\lim}h(f(x)) = h(x_o, \underset{x \to x_o}{\lim}g(x))$

$\displaystyle \displaystyle h(x,c) = \frac{c+A(x)}{2L(x)^5}$

7. Thanks,

I suppose the reason for the inconsistency is because of how FAST the functions approach their limits. You know how 2 functions may intersect at a few points but then not be equal at other points. It is the same with limits. Two functions may have the same limit at Xo but then fail to have the same limit at another x.
But we are evaluating the limit at a specific number x0 not any other point!
Infact the values or the limits of the two functions f and g in other points has nothing to do with the ones at x0.

As for when you can do the substitution that you showed, I think that you can make the procedure more rigorous
Do you mean by substitution we make the limit rigorous?!
1-If yes, it's not always true. Because many complicated limits get easier after doing it...
2-Even if the limit bacomes more complicated, it's not important. We want to know why can not we do it and i think its beeing regorous is not a good reason!
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So what now?

8. Ackbeet's post contains a great example of how the limits DO depend on the neighborhood around x and how fast the functions are going. I suggest you study the example! To avoid counterexamples like that you need to make the procedure rigorous, meaning you have to base it on solid theorems that make it work. With these theorems you can set up the conditions for when it will work.

By substitution i mean using g(x) instead of f(x).

9. I mentioned at first that i know the answer by counterexamples and therefore asked for its reason not by example!
I explained that how by limit laws we get what i got and you gave no mathematical reason for its being false in general except a specific example and a defense by words.

And of course defending of claims only by words has no value in mathematics!

10. I mentioned why! Counterexamples usually lack continuity at the given limit and you cannot perform the following

$\displaystyle \displaystyle \underset{x \to z_o}{\lim} \frac{f(x)}{A(x)} = \frac{\underset{x \to z_o}{\lim}f(x)}{\underset{x \to z_o}{\lim}A(x)}$

Whenever you can perform the following you should be able to exchange f(x) and g(x) because then it is immaterial which function you are using as long as the limit is the same.

Contrary, in counterexamples, you cannot perform the above operation.