How to prove this limit problem?

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In summary: This is because we require f to be differentiable for h > 0, so the restriction on h is already there.
  • #1
Windows_xp
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Homework Statement


Let f : (a, b) → R be a differentiable function and x0 ∈ (a, b). For any h > 0 small, there
exists θ ∈ (0, 1), depending on h, such that
f(x0 + h) = f(x0) + hf'(x0 + θh).

If f is twice differentiable at x0 with f''(x0) != 0, prove that
(a) lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2=(1/2)*f''(x0)
(b) lim h→0 θ =1/2

Homework Equations


lim h→0 (f(c+h)-f(c))/h = f'(c)
L'Hopital's rule

The Attempt at a Solution


For (a),
lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2
=lim h→0 (hf'(x0 + θh)-h*f'(x0))/(h2)
=lim h→0 (f'(x0+θh)-f'(x0))/(2h)
I have no idea what I should do in the next step because the denominator is not θh.Or did I make any mistake in my steps?
 
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  • #2
Windows_xp said:

Homework Statement


Let f : (a, b) → R be a differentiable function and x0 ∈ (a, b). For any h > 0 small, there
exists θ ∈ (0, 1), depending on h, such that
f(x0 + h) = f(x0) + hf'(x0 + θh).

If f is twice differentiable at x0 with f''(x0) != 0, prove that
(a) lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2=(1/2)*f''(x0)
(b) lim h→0 θ =1/2

Homework Equations


lim h→0 (f(c+h)-f(c))/h = f'(c)
L'Hopital's rule

The Attempt at a Solution


For (a),
lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2
=lim h→0 (hf'(x0 + θh)-h*f'(x0))/(h2)
=lim h→0 (f'(x0+θh)-f'(x0))/(2h)
I have no idea what I should do in the next step because the denominator is not θh.Or did I make any mistake in my steps?

If you're allowed (and it looks like you are) I would use L'Hopital for (a) and then use that result from along with work similar to that which you've done to get (b).

There is a slight error in you work; the last line should be $$\lim_{h\rightarrow 0}\frac{f'(x_0+\theta h)-f'(x_0)}{h}.$$ I'm not sure how you got the 2 in the denominator.

Also note that, since ##\theta## is nonzero, $$\frac{f'(x_0+\theta h)-f'(x_0)}{h}=\theta\cdot\frac{f'(x_0+\theta h)-f'(x_0)}{\theta h}.$$

Edit: L'Hopital is not needed for (a); you could get the job done using a fairly straightforward ##\epsilon##-##\delta## argument. But it's much easier to just use l'Hopital if it's available. You might need to justify why you're allowed to use it (i.e. saying more than just 0/0) depending on how much of a stickler the grader is for that kind of thing.

Edit 2: The work that you've done is useful, but I would use it for part (b) rather than part (a). In my opinion, part (a) is more easily done without using the fact that ##f(x_0+h)=f(x_0)+f'(x_0+\theta h)h##.
 
Last edited:
  • #3
gopher_p said:
If you're allowed (and it looks like you are) I would use L'Hopital for (a) and then use that result from along with work similar to that which you've done to get (b).

There is a slight error in you work; the last line should be $$\lim_{h\rightarrow 0}\frac{f'(x_0+\theta h)-f'(x_0)}{h}.$$ I'm not sure how you got the 2 in the denominator.

Also note that, since ##\theta## is nonzero, $$\frac{f'(x_0+\theta h)-f'(x_0)}{h}=\theta\cdot\frac{f'(x_0+\theta h)-f'(x_0)}{\theta h}.$$

Edit: L'Hopital is not needed for (a); you could get the job done using a fairly straightforward ##\epsilon##-##\delta## argument. But it's much easier to just use l'Hopital if it's available. You might need to justify why you're allowed to use it (i.e. saying more than just 0/0) depending on how much of a stickler the grader is for that kind of thing.

The problem with l'hopital is that the numerator is [itex]f'(x_0 + \theta(h)h) - f'(x_0)[/itex] and we don't know that [itex]\theta[/itex] is differentiable in some punctured open neighbourood of [itex]0[/itex].
 
  • #4
pasmith said:
The problem with l'hopital is that the numerator is [itex]f'(x_0 + \theta(h)h) - f'(x_0)[/itex] and we don't know that [itex]\theta[/itex] is differentiable in some punctured open neighbourood of [itex]0[/itex].

Sorry for the confusion. I intend for l'Hopital to be applied immediately to $$\lim_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)-f'(x_0)h}{h^2}$$ without any algebraic rearrangement or use of the fact that ##f(x_0+h)-f(x_0)=hf'(x_0+\theta h)##. The only difficulty here is showing that ##g(h)=f(x_0+h)## is differentiable in a neighborhood of ##0##. It's true, but it's not trivial. I would put it somewhere in the realm of not-hard-to-show-but-worth-mentioning.
 
  • #5
Windows_xp said:

Homework Statement


Let f : (a, b) → R be a differentiable function and x0 ∈ (a, b). For any h > 0 small, there
exists θ ∈ (0, 1), depending on h, such that
f(x0 + h) = f(x0) + hf'(x0 + θh).

If f is twice differentiable at x0 with f''(x0) != 0, prove that
(a) lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2=(1/2)*f''(x0)
(b) lim h→0 θ =1/2

Homework Equations


lim h→0 (f(c+h)-f(c))/h = f'(c)
L'Hopital's rule

The Attempt at a Solution


For (a),
lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2
=lim h→0 (hf'(x0 + θh)-h*f'(x0))/(h2)
=lim h→0 (f'(x0+θh)-f'(x0))/(2h)
I have no idea what I should do in the next step because the denominator is not θh.Or did I make any mistake in my steps?

For part (a), you should try l'hopital's rule directly for the left side of the equation. For part (b), you can use your first attempt's last line and do the trick: multiply by [itex]\theta / \theta [/itex] like gopher_p mentioned and use the fact that [itex]f''(x_0) \neq 0[/itex] to show (b) (you need this fact because [itex]\theta[/itex] is a function of [itex]h[/itex]).

gopher_p said:
Sorry for the confusion. I intend for l'Hopital to be applied immediately to $$\lim_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)-f'(x_0)h}{h^2}$$ without any algebraic rearrangement or use of the fact that ##f(x_0+h)-f(x_0)=hf'(x_0+\theta h)##. The only difficulty here is showing that ##g(h)=f(x_0+h)## is differentiable in a neighborhood of ##0##. It's true, but it's not trivial. I would put it somewhere in the realm of not-hard-to-show-but-worth-mentioning.

It is a trivial application of the chain rule with a certain restriction on [itex]h[/itex] (exercise).
 
  • #6
Actually, I'm not even sure if (b) can be proven with the info given because proving that the limit of [itex]\theta[/itex] as [itex]h[/itex] goes to 0 might be a problem. You can at least say: if the limit of [itex]\theta[/itex] exists, it is 1/2.
 

Related to How to prove this limit problem?

1. How do I know if a limit exists?

To prove that a limit exists, you must show that the left-hand and right-hand limits are equal. If they are not, then the limit does not exist. You can also use the squeeze theorem or epsilon-delta definition to prove the existence of a limit.

2. What is the epsilon-delta definition of a limit?

The epsilon-delta definition of a limit is a formal way to prove the existence of a limit. It states that for every epsilon (ε) greater than 0, there exists a delta (δ) greater than 0 such that if the distance between x and a is less than delta (|x-a|<δ), then the distance between f(x) and L is less than epsilon (|f(x)-L|<ε).

3. How do I use the squeeze theorem to prove a limit?

The squeeze theorem states that if f(x) is always between g(x) and h(x) near a certain point (except possibly at the point itself), and if g(x) and h(x) both have a limit of L as x approaches a, then f(x) must also have a limit of L as x approaches a. To use this theorem, you must find two functions, g(x) and h(x), that are simpler than f(x) and whose limits can be easily proven.

4. Can I use algebraic manipulation to prove a limit?

Yes, algebraic manipulation can be used to prove limits, but it is not always the most efficient method. It is important to remember that when proving a limit, you are showing that the function approaches a certain value as x approaches a, not that the value of the function at a is equal to that value. So, algebraic manipulation may not always be helpful in proving a limit.

5. What are some common mistakes to avoid when proving limits?

Some common mistakes to avoid when proving limits include assuming the existence of a limit without proper proof, using incorrect algebraic manipulations, and mistaking the epsilon-delta definition for the actual value of the limit. It is important to carefully follow the definition and theorems when proving a limit and to double check all steps to avoid errors.

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