L'Hopital's Rule: Advanced Analysis

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Discussion Overview

The discussion revolves around the application of L'Hôpital's Rule in the context of limits involving second derivatives. Participants explore the conditions under which a limit may exist even if the second derivative does not, using specific examples to illustrate their points.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a limit definition involving the second derivative and seeks an example where the limit exists but the second derivative does not.
  • Another participant suggests using the function |x|, which is not differentiable at x=0, as a potential example.
  • Some participants express difficulty in finding a suitable example, noting that attempts with |x| and √x lead to indeterminate forms like 0/0 or infinity when applying L'Hôpital's Rule.
  • There is a challenge regarding whether the limit exists for the proposed functions, with some participants questioning the validity of their calculations.
  • A later reply acknowledges a mistake in the reasoning and seeks clarification on the specific function being discussed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of the limit for the proposed functions, and there are competing views on the applicability of L'Hôpital's Rule in this context.

Contextual Notes

Participants express uncertainty regarding the conditions under which the limit exists and the implications of the non-existence of the second derivative. There are unresolved mathematical steps and assumptions about the functions being analyzed.

Who May Find This Useful

Readers interested in advanced calculus, particularly those exploring the nuances of derivatives and limits, may find this discussion relevant.

TaylorWatts
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Suppose f is defined in a neighborhood of x, and suppose f '' (x) exists. Show that:

lim [f(x+h)+f(x-h)-2f(x)] / h^2 = f''(x).
h->0

Show by an example that that the limit may exist even if f '' (x) may not. (hint: use lHopital's Theorem).

Proof:

f '' (x) exists implies f ' (x) exists and is differentiable which implies f(x) exists and is differentiable.

Thus,

f '' (x) =

lim [f ' (x) - f ' (t)] / (x-t) (def of derivative)
x->t

Doing a substitution h = x - t

lim [f ' (x+h) - f ' (x)] / h (def of derivative)
h->0

= lim [f(x+h) - f(x) - (f(x) - f(x-h)] / h / h
h->0

= lim [f(x+h) + f(x-h) - 2f(x)] / h^2 (*)
h->0

QED

What I'm having trouble with is the example of where the limit exists but f '' (x) does not. Either using the original limit/derivative definition or the end of the proof (*) I get 0/0 or another indeterminate form.
 
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The very simplest function I know that does not have a derivative at one point is |x|, which is not differentiable at x=0. The anti-derivative of that, i.e. f(x) such that f'(x)= |x|, does not have a second derivative at x= 0. Will that work?
 
HallsofIvy said:
The very simplest function I know that does not have a derivative at one point is |x|, which is not differentiable at x=0. The anti-derivative of that, i.e. f(x) such that f'(x)= |x|, does not have a second derivative at x= 0. Will that work?

I tried both absolute value of x and square root of x and still get 0/0 or if I apply l'Hopital's Rule I get infinity. Which also means the limit does not exist.
 
So you have not tried what I suggested? Why not?
 
HallsofIvy said:
So you have not tried what I suggested? Why not?

I did.

The problem is if f ' (x) = |x| then f '' (0) does not exist, but neither does:

lim [f (0 + h) + f (0 - h) - 2 f (0)] / h^2, unless I am missing something here?
h->0
 
TaylorWatts said:
I did.

The problem is if f ' (x) = |x| then f '' (0) does not exist, but neither does:

lim [f (0 + h) + f (0 - h) - 2 f (0)] / h^2, unless I am missing something here?
h->0

That limit does indeed exist. What is f(x) in this case?
 
HallsofIvy said:
That limit does indeed exist. What is f(x) in this case?

I see my mistake.

Thanks!
 

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