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Why don't the derivatives of the functions in question have to be continuous at the point your finding the limit at? Because, if the derivative "jumps", then the limit doesn't equal the value right?

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closed as unclear what you're asking by user7530, AWertheim, hardmath, Sami Ben Romdhane, Daniel Rust Dec 23 '13 at 9:38

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Not sure I understand your question. Is it about L'Hospital's rule? Are you asking whether the rule applies if the derivative is discontinuous at the point? Or what? –  Betty Mock Dec 12 '13 at 3:50
    
I think the question is clear. l'Hospital's rule says that under certain conditions, $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$. One of those conditions is that both $f'$ and $g'$ must be exist. OP wants to know why it works even when $f'$ and $g'$ are discontinuous. –  MJD Dec 12 '13 at 4:22
    
@dfg You said “if the derivative "jumps", then the limit doesn't equal the value right?”. Do you have an example in mind? Because I cannot think of one where this occurs. –  MJD Dec 12 '13 at 4:23
    
To dfg: Do you know a proof for L'Hospital's Rule? In case not, then why do you want continuity of $f'$ and $g'$. If they were continuous the final limit answer would be $f'(a)/g'(a)$. The fact is that the limit is not necessarily $f'(a)/g'(a)$. In fact it may happen that either or both of $f'(a)$ and $g'(a)$ don't exist. A proof of L'Hospital is given at paramanands.blogspot.com/2013/11/… –  Paramanand Singh Dec 12 '13 at 5:07
    
To add to my previous comment, if you are really concerned about jumps of the derivative then no need to worry. A derivative can't have jumps as shown at math.stackexchange.com/a/577978 –  Paramanand Singh Dec 12 '13 at 5:14

1 Answer 1

up vote 3 down vote accepted

There seems to be some confusion about what L'Hopital actual states. Please see, for example, http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_form

In particular, the rule applies to the case where $f,g$ are differentiable on $(a,b)$ and you are evaluating the one-sided limits $\lim_{x \to a^+}$.

To give an example: let $f(x) = |x|$ and $g(x) = x$. Observe that $g$ is differentiable on the whole real line, but $f$ is only differentiable on $(-\infty,0)\cup(0,\infty)$. The rule of L'Hopital thus only applies separately to the two intervals $(-\infty,0)$ and $(0,\infty)$. We have that

$$ -1 = \frac{-x}{x} = \lim_{x \to 0^-} \frac{f(x)}{g(x)} = \lim_{x\to 0^-} \frac{f'(x)}{g'(x)} = \lim_{x\to 0^{-}} \frac{-1}{1} = -1 $$

and

$$ 1 = \frac{x}{x} = \lim_{x\to 0^+} \frac{f(x)}{g(x)} = \lim_{x \to 0^+} \frac{f'(x)}{g'(x)} = \lim_{x\to 0^+} \frac{1}{1} = 1 $$

But evidently the limits

$$ \lim_{x\to 0} \frac{f(x)}{g(x)} \text{ and } \lim_{x \to 0} \frac{f'(0)}{g'(0)} $$

do not exist!


L'Hopital's rule is a statement concerning the equality of the limits of two functions. In a first course of calculus it is, however, often applied in the case where $f$ and $g$ are continuously differentiable in a neighborhood of the point concerned, and so students often end up being used to a statement of the form:

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

when the ratio $f(c)/g(c)$ is indeterminate. Please observe, however, that in the case where $f$ and $g$ are differentiable at $c$ and $f(c) = g(c) = 0$, the proof of L'Hopital's rule becomes trivial: the definition of differentiability states that

$$ f(x) = f'(c)(x-c) + o(|x-c|) \qquad g(x) = g'(c)(x-c) + o(|x-c|) $$ so clearly

$$ \lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(c) + o(1)}{g'(c) + o(1)} = \frac{f'(c)}{g'(c)} $$

In particular, once differentiability is assumed at the limit point $c$, whether $f$ and $g$ are differentiable on a full neighborhood of $c$ is inconsequential. The power of L'Hopital's rule is precisely in the case where $f$ and $g$ are not assumed to be differentiable at the point $c$.


Lastly, what if we are to ask about the case where $f$ and $g$ are differentiable at $c$, but we want to compute the limit using the "limit of the ratio of the derivatives" instead of directly evaluating the derivatives at $c$? Are the two approaches given above guaranteed to yield the same answer? The answer is indeed yes due to Darboux's theorem, which says that if $f$ is a differentiable function (not necessarily continuously differentiable) on an open interval that contains the closed interval $[a,b]$. Then for every $\lambda$ that is between $f'(a)$ and $f'(b)$, there exists $c\in [a,b]$ such that $f'(c) = \lambda$. This in particular implies that "jump discontinuities" are not allowed on an interval over which a function is differentiable.

(Incidentally, the typical differentiable but not continuously differentiable function looks more like $x^2 \cos(1/x)$. Its derivative is $$ \begin{cases} 2x \cos(1/x) + \sin(1/x) & x\neq 0\\ 0 & x = 0\end{cases} $$ When a function is differentiable but not continuously differentiable, the problem is that it oscillates to much, so that at least one of the one-sided limits $\lim_{x\to c^\pm} f'(x)$ do not exist.)

The case where the discontinuity is "essential" (as described in the parenthetical paragraph above) is ruled out, in the statement of L'Hopital's rule, by the assumption that $\lim_{x\to c^\pm} \frac{f'(x)}{g'(x)}$ exists as an extended real number.

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Wow, thanks. Really appreciate the detail. –  dfg Dec 12 '13 at 19:03

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