I was bored and tried to think up a proof for L'hopital's rule in the cas of 0/0.(adsbygoogle = window.adsbygoogle || []).push({});

So, l'hopital's rule states [tex] \lim_{x\rightarrow a} \frac{f(x)}{g(x)} \rightarrow 0/0 [/tex], evaluate the limit using [tex] \frac{f '(x)}{g'(x)} [/tex], 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, [tex] \lim_{x\rightarrow a} = f(a) [/tex]. And because these limits approach 0/0 , we know f(a) and g(a) both equal zero. So:

[itex] \lim_{x\rightarrow a} \frac{f(x) - f(a)}{g(x) -g(a)} \rightarrow 0/0 [/itex]

is identical, as the two subtracted entities are zero. Now, I will multiply by [tex] \frac{x-a} {x-a} [/tex] which is 1. Notice that

[itex] \lim_{x\rightarrow a} \frac{f(x) - f(a)}{(x-a)} [/itex] 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, [tex] F^{-1} [/tex] is a function.

Proof:

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

Proof:

As was shown above, [tex] F^{-1} [/tex] is a function. Furthermore, it was shown that each x matches distinctly to one y. Thus, each y matches to one x. So, [tex] (y_1,x) , (y_2,x) [/tex] implies [tex] y_1 = y_2 [/tex]. 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.

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# Homework Help: A few proof questions

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