Expressions for the second derivative

Suppose that $f$ has continuous second derivatives. How do I show that

$$\frac{f(x+h) + f(x-h) - 2f(x)}{h^2}$$

and

$$2\frac{f(x+h) - f(x) - f'(x)h}{h^2}$$

both tend to $f''(x)$ as $h \rightarrow 0$?

For the first expression, I can rewrite it as

$$\lim_{h \rightarrow 0} \frac{1}{h}(\frac{f(x+h) - f(x)}{h} - \frac{f(x) - f(x-h)}{h})$$

which I can sort of see should tend to $f''(x)$, but I can't seem to show it rigorously. For the second expression, I can rewrite it as

$$\lim_{h\rightarrow 0} \frac{(f(x+h)-f(x))/h - f'(x)}{h}$$

I'm not sure where the factor of 2 comes in, but I guess it should have to do with the fact that we're trying to take limits "simultaneously" for $f'$ and $f''$. Can anyone help? Thanks.

Solution 1:

Apply L'Hospital's rule to differentiate numerator and denominator with respect to $h$: $$ \begin{align*} \lim_{h \to 0} \frac{f(x+h) + f(x-h) - 2f(x)}{h^2} &= \lim_{h \to 0} \frac{f'(x+h) - f'(x-h)}{2h} = f''(x). \end{align*} $$ (Add and subtract $f'(x)$ in the numerator to see the second equality.) Similarly, $$ \begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x) - f'(x)h}{h^2}\, & = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{2h} = \frac{1}{2} f''(x). \end{align*} $$