# A few proof questions

1. Jul 7, 2006

### EbolaPox

I was bored and tried to think up a proof for L'hopital's rule in the cas of 0/0.

So, l'hopital's rule states $$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} \rightarrow 0/0$$, evaluate the limit using $$\frac{f '(x)}{g'(x)}$$, where f and g are differentiable functions.

Proof:
I first noted that f(a) = g(a) = 0, as F and G are differentiable, and thus continuous. So, $$\lim_{x\rightarrow a} = f(a)$$. And because these limits approach 0/0 , we know f(a) and g(a) both equal zero. So:
$\lim_{x\rightarrow a} \frac{f(x) - f(a)}{g(x) -g(a)} \rightarrow 0/0$

is identical, as the two subtracted entities are zero. Now, I will multiply by $$\frac{x-a} {x-a}$$ which is 1. Notice that
$\lim_{x\rightarrow a} \frac{f(x) - f(a)}{(x-a)}$ is the same as the derivative of F evaluated at a. Similarly for g. So, the original limit is f '(a) / g'(a).

This sounds poorly written, but that's all I have so far. Any major errors in it that need to be corrected? It made sense to me when I wrote it.

Another proof:

If F is a one-to-one function, then the inverse of F, $$F^{-1}$$ is a function.

Proof:
F being a function indicates that there are sets of ordered pairs (x,y). Because F is a function, $$(x,y_1) , (x,y_2)$$ indicates $$y_1 = y_2$$, as a function matches one x value to one y value. Next, because we know F is one-to-one, $$(x_1, y) , (x_2 ,y)$$ implies $$x_1 = x_2$$. So, each distinct x is matched to one distinct y. Now, $$F^{-1}$$ is the set of ordered pairs (y,x). For $$F^{-1}$$ to be a function, then each y must match distinctly to one x, or in other words,$$(y, x_1) , (y,x_2)$$ implies $$x_1 = x_2$$. This must be so, as it was shown above that each x matches distinctly to one y.

An addition to that one:
If F is one to one , then $$F^{-1}$$ is also one-to-one.

Proof:
As was shown above, $$F^{-1}$$ is a function. Furthermore, it was shown that each x matches distinctly to one y. Thus, each y matches to one x. So, $$(y_1,x) , (y_2,x)$$ implies $$y_1 = y_2$$. This is true as each distinct y mathces to a distinct x.

I hope my notation makes sense and the LaTeX came out alright. I'm kinda new to proof writing (my high school didn't ever ask me to prove anything.)

Thanks for any suggestions/corrections/hints!

Edit: Had to fix all of my latex formulas which came out odd looking.

Last edited: Jul 7, 2006
2. Jul 7, 2006

### StatusX

First, remember that in general:

$$\frac{\lim_{x \rightarrow a} A(x) }{ \lim_{x \rightarrow a} B(x) } \neq \lim_{x \rightarrow a} \frac{A(x)}{B(x)}$$

This only holds if both limits on the LHS exist, and if the denominator of the LHS is non-zero. You need to use the fact that f(x) and g(x) are differentiable at x=a to get past this step in your proof, and you'll need to treat the case where the denominator of the LHS is zero seperately. Which brings me to the other problem I see, which is that you should have:

$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$

Since you may again have f'(x)=g'(x)=0 or infinity. The second proof looks OK.

3. Jul 8, 2006

### matt grime

No, that is false. Bijective functions are the only ones to have inverse, not injective ones.

4. Jul 8, 2006

### EbolaPox

Thanks for the replies. I'll think up something to fix my L'hopital error soon.

With respect to the inverse function proof, I quoted the theorem straight out of a textbook and tried to prove it. Should I instead attempt to prove "If F is a one-to-one and onto function, then $$F^{-1}$$ is a function".

I see exactly what you're saying, I recall reading such in my Linear Algebra text. I'm not quite sure why the other book I'mr eading had an erroneous theorem in it. Thank you for pointing that out. I shall correct these proofs soon.

5. Jul 8, 2006

### StatusX

Any injective function can trivially be made bijective by restricting the range to the image of the domain. They probably mean that the function has an inverse on this latter set.

6. Jul 8, 2006

### matt grime

f^{-1} as a function cannot exist unless f is both injective and surjective. f^{-1} certailny can be defined set wise for any function f, i.e. f^{-1}(U) is the set { v | f(v) is in U}. It is perfectly acceptable to have this inverse image the empty set.

This can be used to define a function if and only if f is bijective. You cannot define f^{-1}(u) for any u not in the range of f otherwise.