L'Hopital's Rule: Advanced Analysis

In summary, the conversation discusses how to show that if f '' (x) exists, then the limit of [f(x+h)+f(x-h)-2f(x)] / h^2 is equal to f''(x). An example using l'Hopital's Theorem is also provided to show that the limit may exist even if f '' (x) does not. The example given is the function |x|, which is not differentiable at x=0 and therefore does not have a second derivative at that point. However, the limit still exists.
  • #1
TaylorWatts
16
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
So you have not tried what I suggested? Why not?
 
  • #5
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
 
  • #6
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?
 
  • #7
HallsofIvy said:
That limit does indeed exist. What is f(x) in this case?

I see my mistake.

Thanks!
 

FAQ: L'Hopital's Rule: Advanced Analysis

What is L'Hopital's Rule and when is it used?

L'Hopital's Rule is a mathematical tool used to evaluate limits of indeterminate forms, where both the numerator and denominator approach zero or infinity. It is typically used in advanced analysis and calculus to simplify complex functions.

How does L'Hopital's Rule work?

L'Hopital's Rule is based on the concept that the ratio of the derivatives of two functions is equivalent to the ratio of the original functions, as long as the limit exists. This allows for the evaluation of indeterminate forms by taking the derivative of the numerator and denominator separately and then evaluating the limit again.

What are some common indeterminate forms that can be evaluated using L'Hopital's Rule?

Some common indeterminate forms include 0/0, ∞/∞, 0 * ∞, ∞ - ∞, 1^∞, and ∞^0. These forms can arise from the evaluation of limits or when solving certain types of integrals.

Are there any restrictions or limitations to using L'Hopital's Rule?

There are a few restrictions to using L'Hopital's Rule. First, the limit must be in an indeterminate form in order for the rule to be applied. Additionally, both the numerator and denominator must be differentiable functions and the limit must exist. Lastly, the use of L'Hopital's Rule should be limited to situations where it simplifies the problem or makes it easier to solve, rather than using it as a shortcut for every limit problem.

Can L'Hopital's Rule be used to evaluate limits at infinity?

Yes, L'Hopital's Rule can be used to evaluate limits at infinity. In this case, the numerator and denominator of the original function can be rewritten as a polynomial in terms of 1/x, and then the rule can be applied as usual. However, it is important to note that this only works if the limit at infinity is in an indeterminate form.

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