# Prove that the following limit equals $f''(a)$ if $f''(a)$ exists

$$1.$$ Suppose that $$f''(a)$$ exists. Show that $$\lim\limits_{h\to 0} \dfrac{f(a+h)+f(a-h)-2f(a)}{h^2}=f''(a).$$

$$2.$$ Show by example that this limit may exist even when $$f''(a)$$ does not.

My work:

$$1.$$

By the derivative definition, $$f''(a) = \lim\limits_{h\to 0}\dfrac{f'(a+h)-f'(a)}{h}\\$$ $$=\lim\limits_{h\to 0} \dfrac{f'(a)-f'(a-h)}{h}.$$ To see this, let $$k=-h.$$ Then $$k\to0\Leftrightarrow h\to 0$$ and $$\lim\limits_{h\to 0}\dfrac{f'(a)-f'(a-h)}{h} = \lim\limits_{k\to 0}\dfrac{f'(a)-f'(a+k)}{-k}\\ =\lim\limits_{k\to 0}\dfrac{f'(a+k)-f'(a)}{k}\\ =\lim\limits_{h\to 0}\dfrac{f'(a+h)-f'(a)}{h}.$$ So the limit is equivalent to $$\lim\limits_{h\to 0}\dfrac{\frac{f(a+h)-f(a)}{h}-\frac{f(a)-f(a-h)}{h}}{h}\\ =\lim\limits_{h\to 0}\dfrac{f(a+h)+f(a-h)-2f(a)}{h^2}.$$

$$2.$$

Consider $$f(x)=\begin{cases} x^2\sin (1/x)& x\neq 0\\ 0& x=0\end{cases}.$$

We have that $$f'(x) = 2x\sin(1/x)-\cos(1/x),x\neq 0$$ and $$f''(x) = 2\sin(1/x)-\dfrac{2}{x}\cos (1/x)-\dfrac{\sin(1/x)}{x^2},x\neq 0.$$ Note that $$f'(0)=\lim\limits_{h\to 0}\dfrac{h^2\sin (1/h)}{h}\\ =\lim\limits_{h\to 0} h\sin (1/h).$$

Also, note that $$\forall h>0, -h\leq h\sin(1/h)\leq h$$ and $$\forall h\leq 0,h \leq h\sin (1/h)\leq -h.$$ Hence by the Squeeze Theorem, $$\lim\limits_{h\to 0}h\sin (1/h)=\lim\limits_{h\to 0}h = 0.$$ In order for $$f''(0)$$ to exist, we must have that $$f'(x)$$ is differentiable at $$x=0.$$ However, we will show that $$f'(x)$$ is discontinuous at $$x=0$$ and hence not differentiable there. We will do so by showing that $$\lim\limits_{x\to 0^-}f'(x)$$ does not exist. Consider the sequence $$(x_n)_{n=1}^\infty$$ such that $$x_n = -\dfrac{1}{\frac\pi2 + 2n\pi}$$ and the sequence $$(y_n)_{n=1}^\infty$$ such that $$y_n=-\dfrac{1}{\frac{3\pi}{2}+2n\pi}.$$ $$\lim\limits_{x\to 0^-}f'(x)$$ does not exist because $$x_n, y_n\to 0$$ as $$n\to \infty\Rightarrow \forall \epsilon>0, \exists N (n\geq N \Rightarrow x_n,y_n \in (-\epsilon,0)).$$ Since $$f'(x_n)<0 we have that $$f''(0)$$ does not exist.

However, we have that

$$\lim\limits_{h\to 0}\dfrac{f(0+h)+f(0-h)-2f(0)}{h^2}=\lim\limits_{h\to 0}\dfrac{h^2\sin (1/h)-h^2\sin(1/h)}{h^2}\\ =0.$$ Thus, the limit exists at $$x=0$$ but the second derivative does not.

edit for the first part (i should've used the taylor series instead).

We have that $$f(a+h) = f(a) + f'(a)h+f''(a)\dfrac{h^2}{2}+\dots$$ and $$f(a-h)=f(a)-f'(a)h+f''(a)\dfrac{h^2}{2}+\dots.$$ Hence $$f(a+h)+f(a-h)-2f(a)=h^2f''(a)$$ and the desired limit is $$\lim\limits_{h\to 0} \dfrac{h^2f''(a)}{h^2}=f''(a),$$ as desired.

• How are you concluding that $f''(0)$ does not exist? – Daniel Schepler Nov 8 '19 at 22:59
• I am concluding that because it is undefined at $x=0.$ However, its limit as $x\to 0$ is indeed defined. Is there something wrong? – user718615 Nov 8 '19 at 23:00
• Because what is undefined at $x=0$? (If you're referring to the expression for $f''(x)$ for $x \ne 0$, that's not sufficient. e.g. $f'(0)$ exists and in fact $f'(0) = 0$, even though the expression for $f'(x)$ for $x \ne 0$ is undefined at $x=0$.) – Daniel Schepler Nov 8 '19 at 23:02
• (Hint for a valid proof to show that $f''(0)$ does not exist: if you can show that $f'$ is not continuous at $x=0$, that implies that $f'$ is not differentiable at $x=0$.) – Daniel Schepler Nov 8 '19 at 23:07
• @DanielSchepler thanks. – user718615 Nov 8 '19 at 23:07

Your approach for first part has subtle mistakes. You can't use the same variable $$h$$ in two different limit context and then treat them as same.

Your approach should lead to the expression $$\lim_{h\to 0}\frac{f'(a+h)-f'(a)}{h}=\lim_{h\to 0}\dfrac{\lim\limits _{k\to 0}\dfrac{f(a+h)-f(a+h-k)}{k}-\lim\limits _{k\to 0}\dfrac{f(a+k)-f(a)}{k}}{h}$$ and those two limit variables $$h, k$$ are different and can't be combined by writing $$h=k$$.

The right approach is to use L'Hospital's Rule or Taylor series. Via L'Hospital's Rule the expression in question is reduced to $$\frac{f'(a+h) - f'(a-h)} {2h}$$ and this tends to $$f''(a)$$ via obvious algebraic manipulation.

Your approach for second part is fine.

• but what if I want to solve the problem without using the Taylor series for $f'(a+h)$ and $f'(a-h)$? – user718615 Nov 9 '19 at 1:27
• @FredJefferson: I mentioned algebraic manipulation and not Taylor at this step. Add and subtract $f'(a)$ in numerator and split like $$\frac{f'(a+h) - f'(a)} {2h}+\frac{f'(a)-f'(a-h)}{2h}$$ and each of the fractions above tends to $f''(a) /2$. – Paramanand Singh Nov 9 '19 at 2:02
• But I still don't understand why we can use L'Hôpital's rule here. L'Hôpital's rule has a lot of conditions, as I've seen in a previous problem. – user718615 Nov 9 '19 at 2:05
• @FredJefferson: all the conditions of the rule are met here. The numerator and denominator of the expression are functions of $h$ which tend to $0$ as $h\to 0$. Further the expression after applying L'Hospital's Rule tends to a specific limit (as i have shown in previous comment). Hence by L'Hospital's Rule the original expression also tends to the same limit. Do you think that there are some conditions which are not met for L'Hospital Rule? – Paramanand Singh Nov 9 '19 at 2:12
• @FredJefferson: also note that for this problem you can't apply L'Hospital's Rule twice because then the conditions are not met. – Paramanand Singh Nov 9 '19 at 2:14

Your solution seems right and fine.

For the first part we can also use Taylor's expansion

• $$f(a+h)=f(a)+f'(a)h+\frac12 f''(a)h^2+o(h^2)$$
• $$f(a-h)=f(a)-f'(a)h+\frac12 f''(a)h^2+o(h^2)$$

therefore

$$\dfrac{f(a+h)+f(a-h)-2f(a)}{h^2}=\frac{f''(a)h^2+o(h^2)}{h^2}=f''(a)+o(1) \to f''(a)$$

• are you sure the second part is correct? $\sin(1/x)$ doesn't even have a limit at $x=0.$ – user718615 Nov 8 '19 at 22:23
• @FredJefferson The expression $$\dfrac{h^2\sin (1/h)-h^2\sin(1/h)}{h^2}=0$$ is identically equal to zero, therefore the limit is zero. – user Nov 8 '19 at 22:26
• how do you know that $f(a+h)=f(a)+f'(a)h+\dots$ and that $f(a-h)=f(a)-f'(a)h+f''(a)\dfrac{h^2}{2}$? Is this some kind of linear approximation? – user718615 Nov 9 '19 at 1:57
• @FredJefferson It is Taylor expansion with Peano's remainder. – user Nov 9 '19 at 2:09