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$1.$ 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 $f'(x_n)<0<f'(y_n)\;\forall n,$ 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.

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2 Answers2

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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.

Paramanand Singh
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  • but what if I want to solve the problem without using the Taylor series for $f'(a+h)$ and $f'(a-h)$? –  Nov 09 '19 at 01:27
  • @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 09 '19 at 02:02
  • 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. –  Nov 09 '19 at 02:05
  • @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 09 '19 at 02:12
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    @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 09 '19 at 02:14
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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

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

user
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  • are you sure the second part is correct? $\sin(1/x)$ doesn't even have a limit at $x=0.$ –  Nov 08 '19 at 22:23
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    @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 08 '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? –  Nov 09 '19 at 01:57
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    @FredJefferson It is Taylor expansion with Peano's remainder. – user Nov 09 '19 at 02:09