# Homework Help: Need help with these Derivatives

1. Jul 29, 2010

### jgens

1. The problem statement, all variables and given/known data

If $f$ is differentiable at $a$, prove the following:
$$\lim_{h,k \to 0^+} \frac{f(a+h)-f(a-k)}{h+k} = f'(a)$$

2. Relevant equations

N/A

3. The attempt at a solution

At the moment, I don't have a complete proof worked out, but I was wondering if someone could comment on the validity of this reasoning . . .

Clearly, for every $\varepsilon > 0$, there exists a $\delta_1 > 0$ such that if $0 < k < \delta_1$ then . . .

$$\left|\frac{f(a+h)-f(a-k)}{h+k} - \frac{f(a+h) - f(a)}{h} \right| < \frac{\varepsilon}{2}$$*​

Moreover, for this same $\varepsilon$, there must be some other number $\delta_2 > 0$ such that whenever $0 < h < \delta_2$, it follows that . . .

$$\left|\frac{f(a+h)-f(a)}{h} - f'(a) \right| < \frac{\varepsilon}{2}$$​

From this, so long as $0 < k < \delta_1$ and $0 < h < \delta_2$, we have

$$\left|\frac{f(a+h)-f(a-k)}{h+k} - f'(a) \right| < \varepsilon$$​

as desired.

*I realize that this is the point that really needs some work, but I think that it should be a trivial (albeit potentially long-winded) exercise to find the proper $\delta$.

2. Jul 30, 2010

### losiu99

Re: Derivatives

Looks fine, just remamber that the first inequality with given $$\delta_1$$ must hold for all $$0 < h <\delta_2$$.

3. Jul 30, 2010

### jgens

Re: Derivatives

Well, the first inequality should hold for any arbitrary h > 0, so I should have my bases covered. Thanks! I'll work on finding a proper $\delta$ now.

4. Jul 30, 2010

### losiu99

Re: Derivatives

Since we don't have any global assumptions on f, I don't think it's possible to find $$\delta$$ for arbitrary h. If, for example, for some h>0 function
$$\phi_h(k)=\frac{f(a+h)-f(a-k)}{h+k}$$
is discontinuous at 0, it's hopeless.

5. Jul 30, 2010

### jgens

Re: Derivatives

Well, I'm working on my phone right now, so I'm not sure if this is correct, but here are my thoughts. For every $\varepsilon > 0$, it's clearly possible to find a $\delta_1 > 0$ such that if $0 < h < \delta_1$, then the second inequality holds. Based on this choice of $h$ and the continuity of $f$ at $a$, it's then possible to choose a $\delta_2 > 0$ such that if $0 < k < \delta_2$, then the second inequality holds. This would show that for every number $\varepsilon > 0$, it's possible to find the necessary two numbers $\delta > 0$ for the final inequality to hold. However, looking through this, the dependence of $\delta_2$ on h seems like it might be problematic. Ugh :(

Edit: I'm fairly certain that it's possible to find the necessary $\delta$s that I allude to in this post; it's just is the second one's reliance on h problematic or is it a non-issue.

6. Jul 30, 2010

### jgens

Re: Derivatives

Thanks! I'll try to post a fully worked out proof later tonight when I have access to a computer. If you don't mind, could you explain why the delta with dependence on h is problematic? I understand why it's certainly problematic if the delta is dependent on the same variable it's placing a bound on, but I can't figure it out exactly when it's dependent on a different variable. Thanks again!

7. Jul 30, 2010

### jgens

Re: Derivatives

I would probably be able to replace delta 2's dependence on h with a dependence on delta 1. I'm not sure if this makes the situation any better, but I could really use some help here. Thanks.

8. Jul 30, 2010

### losiu99

Re: Derivatives

Sorry for deleting that post. Dependence is indeed problematic because of the definition of double limit, but I realised I made a critical mistake in my solution. I'll think about and post if I find anything useful. Good luck!

9. Jul 30, 2010

### Dick

Re: Derivatives

Try this. Can you show min(L1,L2)<=(h*L1+k*L2)/(h+k)<=max(L1,L2) for h and k positive? Can you think how you might use that?

10. Jul 30, 2010

### jgens

Re: Derivatives

Thanks for the help Dick! First, we need only note that ...

$$h\min{(L_1,L_2)} + k\min{(L_1,L_2)} \leq h \dot L_1 + k \dot L_2 \leq h\max{(L_1,L_2)} + k\max{(L_1,L_2)}$$​

From which the inequality that you posted follows immediately. Next, we define $L_1$ and $L_2$ such that ...

$$L_1 = \frac{f(a+h)-f(a)}{h}$$
$$L_2 = \frac{f(a)-f(a-k)}{k}$$

Using these definitions, we have that ...

$$\min{(L_1,L_2)} \leq \frac{f(a+h)-f(a)+f(a)-f(a-k)}{h+k} = \frac{f(a+h)-f(a-k)}{h+k} \leq \max{(L_1,L_2)}$$​

And then using the squeeze theorem or explicitly writing out the $\varepsilon-\delta$ proof, we prove the desired result. Is this what you were getting at Dick?

11. Jul 30, 2010

### Dick

Re: Derivatives

Sure, it's a squeeze theorem. And you know how to apply it. So why is min(L1,L2)<=(h*L1+k*L2)/(h+k)<=max(L1,L2) for h,k>0? Do you get that part?

12. Jul 30, 2010

### jgens

Re: Derivatives

The first part of my post was meant to address that, I'm sorry if it wasn't clear. To outline it again without that pain of the maxs/mins, I'll assume (without loss of generality) that $L_1 \leq L_2$. Clearly ...

$$L_1(h+k) \leq h \dot L_1 + k \dot L_2 \leq L_2(h+k)$$​

Dividing through by $h+k$ we get ...

$$L_1 \leq \frac{h \dot L_1 + k \dot L_2}{h+k} \leq L_2$$​

as desired.

13. Jul 30, 2010

### jgens

Re: Derivatives

Thanks for your help anyway. I'm not familiar at all with the definition of a limit involving two variables (single variable calc. course here) so that makes sense why I was having such a difficult time explicitly understanding why the dependence was so problematic; after looking up the correct definition, it makes complete sense.

14. Jul 30, 2010

### Dick

Re: Derivatives

That's a bit of circular reasoning and why are you putting dots on the L's? I know L1<=(h*L1+k*L2)/(h+k)<=L2 is true for L1<=L2 and h,k>0. But do you know why? I'll give you a hint. It's the same reason why L1<=t*L1+(1-t)*L2<=L2 for t in [0,1].

15. Jul 30, 2010

### jgens

Re: Derivatives

Sorry about the dots, they were supposed to be multiplication signs. And what's particularly circular about my reasoning? If I choose arbitrary $h,k > 0$ and assume that $L_1 \leq L_2$, then $hL_1 + kL_1 \leq hL_1 + kL_2$ since $kL_1 \leq kL_2$. The other inequality would follow from similar reasoning.

I'm off to bed now, so I'll follow up on your last idea tomorrow morning. Thanks again Dick!

16. Jul 30, 2010

### Dick

Re: Derivatives

Oh, nothing circular about your thinking. More of a problem with my following what you are doing. I was focusing on the dots. Don't worry about it.

17. Jul 31, 2010

### losiu99

Re: Derivatives

We can solve it similarly without any min/max reasoning. Observe that
$$\left|\frac{f(a+h)-f(a-k)}{h+k}-f'(a) \right| = \left|\frac{f(a+h)-f(a) + f(a)-f(a-k)}{h+k}-f'(a)\right| =$$
$$\left|\frac{f(a+h)-f(a) -hf'(a)+ f(a)-f(a-k)-kf'(a)}{h+k}\right| \leq$$
$$\left|\frac{f(a+h)-f(a) -hf'(a)}{h+k}+\frac{f(a) - f(a-k)-kf'(a)}{h+k}\right| \leq$$
$$\frac{h}{h+k} \left|\frac{f(a+h)-f(a)}{h} -f'(a)\right| +\frac{k}{h+k} \left|\frac{f(a-k)-f(a)}{-k} -f'(a)\right| \leq$$
$$\left|\frac{f(a+h)-f(a)}{h} -f'(a)\right|+\left|\frac{f(a-k)-f(a)}{-k} -f'(a)\right|$$
Sorry for not posting this earlier, I wanted to complete you original idea.

Last edited: Jul 31, 2010
18. Jul 31, 2010

### jgens

Re: Derivatives

Thanks again Dick! And nice proof losiu99! I really appreciate all the help that you've given me.