# Equations with limits

1. Dec 11, 2013

### Jhenrique

I believed the definitions of derivative that we know was really definitions
$$f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$$
$$f'(x_0)=\lim_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$
But not, is one definition, just use the equality bellow in equations above...
$$\\ \Delta x=x-x_0 \\ x=x_0+\Delta x$$

However, how proof the equality between limits?
$$\lim_{x\rightarrow x_0} \; \underset{=}{?} \; \lim_{\Delta x\rightarrow 0}$$

Maybe by:
$$\lim_{x\rightarrow x_0}=\lim_{x-x_0\rightarrow x_0-x_0}=\lim_{\Delta x\rightarrow 0}$$

What do you think about this? Is correct work with limit as if it is a equation?

EDIT: And more... Is possible to manipulate the limits like a algebric variable? I want say... move the limit from left side of equation to right side, apply a limit in equation that eliminate outher limit that already exist, know, use the limit like a function (and a function have a inverse function). What do you think??

Last edited: Dec 11, 2013
2. Dec 11, 2013

### Stephen Tashi

That notation doesn't make sense. You should specify the functions involved. To properly phrase your question, you should ask:

How do we prove:

[eq, 1] $\lim_{x\rightarrow x_{0}} g(x) = \lim_{\Delta x \rightarrow 0} g(x + \Delta x)$

How you prove things depends on what theorems you have already proven. You seem to be seeking a proof that only uses the manipulation of symbols. Many important mathematical theorems cannot be proven just by manipulation of symbols. (Symbolic manipulation is an aid to thought, not a substitute for it.) One way to prove [eq. 1] is to write a proof consisting mostly of words. Another way is to use theorems about limits of the composition of functions, if your study materials have proven such theorems.

It is not possible in general. For special kinds of functions, you might be able to develop methods of symbolic manipulation that work. For example, in general $lim_{x\rightarrow x_{0}} lim_{y\rightarrow y_{0}} f(x_0,y_0)$ is not equal to $lim_{y\rightarrow y_{0}} lim_{x\rightarrow x_{0}} f(x_0,y_0)$. However, for many types of functions, the two expressions are equal.

3. Dec 11, 2013

### Jhenrique

Given:
$$\frac{f(x)-f(x_0)}{x-x_0}=\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$
and applying one limit bellow in equation above
$$\\ \lim_{x\rightarrow x_0} \\ \lim_{\Delta x\rightarrow 0}$$
then:
$$\lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\rightarrow x_0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$
So this I saw the need of know to transform the limits so that which equation have its apropried limit

4. Dec 11, 2013

### Staff: Mentor

The limit on the right doesn't make sense - there is no x anywhere. That limit should be as Δx → 0.

If Δx = x - x0, then as x → x0, Δx → 0. It might be that that's what you're asking about.

5. Dec 11, 2013

### Jhenrique

Yes, doesn't make sense. Exactly by this that it need to be manipuled: x → x₀ <=> x - x₀ → x₀ - x₀ <=> Δx → 0

6. Dec 11, 2013

### Staff: Mentor

BTW, the word is manipulated

7. Dec 12, 2013

### Jhenrique

So so, I'd like to know the your opinion about this. But, I already noticed that is possible work with limits in a equation.

Sorry I'm not american or british...

8. Dec 15, 2013

### Staff: Mentor

Yes, you can convert limits as you show above. It's nothing more complicated than algebra. If Δx = x - x0, then as x approaches x0, Δx necessarily approaches 0.