How to prove this limit problem?

[Windows_xp]

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?

 

[gopher_p]

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 $f(x_0+h)=f(x_0)+f'(x_0+\theta h)h$.

 

[pasmith]

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 f'(x_0 + \theta(h)h) - f'(x_0) and we don't know that \theta is differentiable in some punctured open neighbourood of 0.

 

[gopher_p]

The problem with l'hopital is that the numerator is f'(x_0 + \theta(h)h) - f'(x_0) and we don't know that \theta is differentiable in some punctured open neighbourood of 0.

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.

 

[exclamationmarkX10]

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 \theta / \theta like gopher_p mentioned and use the fact that f''(x_0) \neq 0 to show (b) (you need this fact because \theta is a function of h).

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 h (exercise).

 

[exclamationmarkX10]

Actually, I'm not even sure if (b) can be proven with the info given because proving that the limit of \theta as h goes to 0 might be a problem. You can at least say: if the limit of \theta exists, it is 1/2.