# Thread: Limit Formula For Second Derivative

1. ## Limit Formula For Second Derivative

Show that $\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

$\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}$

But also that:

$\displaystyle f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}$

$\displaystyle f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

So when I substitute these in, I'm instead getting the following:

$\displaystyle \displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$

$\displaystyle =\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$

Which I know is just off by an $\displaystyle h$. It seems to me that they get the second derivative formula like this:

$\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}$

But I don't really understand why it's that way. That seems like $\displaystyle f''(x-h)$ to me. Why is it done that way?

2. Look here,

3. Originally Posted by mathematicalbagpiper
Show that $\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

$\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}$

But also that:

$\displaystyle f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}$

$\displaystyle f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

So when I substitute these in, I'm instead getting the following:

$\displaystyle \displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$

$\displaystyle =\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$

Which I know is just off by an $\displaystyle h$. It seems to me that they get the second derivative formula like this:

$\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}$

But I don't really understand why it's that way. That seems like $\displaystyle f''(x-h)$ to me. Why is it done that way?
It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)

4. Originally Posted by Also sprach Zarathustra
It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)
You mean x=y+h...

5. Originally Posted by Pandevil1990
You mean x=y+h...

No... x=y-h will work.

6. where exactly are you going to do this substitution?

7. Originally Posted by Pandevil1990
where exactly are you going to do this substitution?
Here:

$\displaystyle \displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$

8. It's ok...But i think this is a more difficult way than L'hospital 's rule.Thanks...