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I was reading Spivak's Calculus. I have a query with regards to the proof for L'Hopital's rule for the 0/0 indeterminate form. Theorem Statement :-

"Suppose that $ \lim_{x\to a} f(x) = 0 $ and $\lim_{x\to a} g(x) = 0 $ and suppose that $ \lim_{x\to a} \frac {f'(x)}{g'(x)} $ exists . Then $ \lim_{x\to a} \frac {f(x)}{g(x)} $ exists and

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

I have a confusion with regards to the last part of the proof.

Using the Cauchy Mean Value Theorem it is shown that there exists a number $\alpha_x$ in $(a,x)$ such that $$ \ \frac {f(x)}{g(x)} = \ \frac {f'(\alpha_x)}{g'(\alpha_x)} $$

Now $\alpha_x$ approaches $a$ as $x$ approaches $a$ because $\alpha_x$ is in $(a,x)$ , it follows that $$ \lim_{x\to a} \frac {f(x)}{g(x)} = \lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{y\to a} \frac {f'(y)}{g'(y)} $$

My query is regarding the last two of the above equations. I understand that as $x$ approaches $a$ so does $\alpha_x$ , but then how are we treating $\alpha_x$ as a 'dummy variable' and replacing it with $y$. Is not $\alpha_x$ a dependent variable on x. (I am also having some trouble in seeing how the step $$\lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)}$$ is justified )

Thanks in advance. I am new to these forums and looking forward to the discussion.

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Thanks everyone for answering . All the answers were most helpful. And my confusion is cleared, I think. –  ameyask86 Oct 28 '12 at 9:37
    
If your problem has been solved, there should be a solution for your question, so please be sure to come back and mark an answer as accepted. –  phunehehe Oct 28 '12 at 10:41
    
Ok, I will mark the answer which explained the convergent sequence approach. As my confusion arose when I attempted it using that approach. But really , all the answers were very helpful. –  ameyask86 Oct 28 '12 at 11:57
    
I can't find the post now, but it is considered good practice to upvote all answers that you find helpful, and accept the one that solves the problem in the best way :) –  phunehehe Oct 28 '12 at 12:25
    
Yes, I upvoted all the answers. :) –  ameyask86 Oct 28 '12 at 13:11

3 Answers 3

up vote 2 down vote accepted

I see your point and it may be viewed as sloppy formulation. We start on the left with an arbitrary sequence $x\to a$ and want to show that we obtain the same value for all such arbitrary sequences. Then we replace this arbitrary sequence with a specific sequence $\alpha_x\to a$ (depending on the given arbitray sequence). Since the premise is that the right hand side limit exists, i.e. that we obtain the same value for all sequences $y\to a$, we know that our specific sequence yields that very value.

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"Then we replace this arbitrary sequence with a specific sequence αx→a (depending on the given arbitray sequence). Since the premise is that the right hand side limit exists, i.e. that we obtain the same value for all sequences y→a, we know that our specific sequence yields that very value." Thanks that really helped. I will think over it and try to paraphrase so as to confirm whether I got it. –  ameyask86 Oct 28 '12 at 8:28
    
Though there is the epsilon-delta approach to find a limit.Let us see if I can try it using the sequences approach, so as to be conceptually clear. Consider an 'arbitrary' sequence {x_n} which converges to 'a' and see if 'f(x_n)/g(x_n)' converges to a real number.By CMVT that f(x)/g(x) = f'(αx)/g'(αx). We know lim x→a (f'(x)/g'(x)) exists. Thus for any sequence converging to 'a' the corresponding f'(x_n)/g'(x_n) sequence converges to l. So this is true even for the specific sequence αx_n→a .Now, as x_n→a , αx_n→a. Thus , f(x_n)/g(x_n) which equals f'(αx_n)/g'(αx_n) also converges to l. –  ameyask86 Oct 28 '12 at 9:27
    
I posted in the previous comment the way one could go about proving it using arbitrary sequence {x_n} . Is my approach right ? Sorry if I caused any inconvenience by posting in multiple comments. –  ameyask86 Oct 28 '12 at 9:32

The last two identities follow from the following fact:

Assume that the function $h$ has limit $\ell$ at $a$ and that the function $\alpha$ has limit $a$ at $x_0$, that is, $$ \lim_{t\to a}h(t)=\ell,\qquad\qquad\lim_{x\to x_0}\alpha(x)=a. $$ Then, the function $h\circ\alpha$ has limit $\ell$ at $x_0$, that is, $$ \lim_{x\to x_0}h(\alpha(x))=\ell. $$

A simple proof is by the usual epsilon-delta approach.

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Thanks. Will use the epsilon-delta approach to prove the result you mentioned about composite functions. And use that in the proof to clear my confusion. –  ameyask86 Oct 28 '12 at 8:29

These equalities hold because you assumed that the limit $\displaystyle\lim_{x\rightarrow a} \dfrac{f'(x)}{g'(x)}$ exists. That means it doesn't really matter how you approach $a$ (i.e. either take $x\rightarrow a$ or $\alpha_x\rightarrow a$), you have to arrive at the same value for the limit. This also explains why your last equality is okay, since you have some variable approaching $a$, it doesn't matter how you get there; essentially if the limit exists, then the variable respect to which you are taking the limit is automatically a dummy variable.

If you want to be more precise, you can use the $\varepsilon$-$\delta$ definition of the limit. Say $L$ is your limit, which we assume exists. Then for all $\varepsilon>0$ there exists some $\delta>0$ such that if $|a-x|<\delta$ then $\left|L-\dfrac{f'(x)}{g'(x)}\right|<\varepsilon$. But $|a-\alpha_x|<|a-x|<\delta$, so we know that for each $\varepsilon$, we can use the same $\delta$ to see that if $|x-a|<\delta$, then $\left|L-\dfrac{f'(\alpha_x)}{g'(\alpha_x)}\right|<\varepsilon$. This immediately implies $\displaystyle \lim_{x\rightarrow a}\dfrac{f'(\alpha_x)}{g'(\alpha_x)} = L$.

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