Prove without using L'Hopital's rule that $f''(a)=0$ if $\lim_{x\to a}{\frac{f(x)-f(a)}{(x-a)^2}}=0$

Question

I'm trying to prove without using L'Hopital's rule that $f''(a)=0$ if

$$\lim_{x\to a}{\frac{f(x)-f(a)}{(x-a)^2}}=0,$$

assuming $f$ is twice differentiable and $f''$ is continuous. My first instinct was to use the mean value theorem to define a real number $t$ such that $f'(t)=\frac{f(x)-f(a)}{x-a}$, which gives $\lim_{x\to a}{\frac{f'(t)}{x-a}}=0$. Since $t\to a$ as $x\to a$, this gives $f'(a)=0$, and

$$\lim_{x\to a}{\frac{f'(t)}{x-a}}=\lim_{x\to a}{\frac{f'(t)}{f'(x)}}\times\lim_{x\to a}{\frac{f'(x)}{x-a}}=1\times f''(a)=0,$$

thus $f''(a)=0$. I wasn't entirely convinced, however, and quickly found counterexamples such as $\lim_{x\to0}{\frac{\sin x^3}{\sin x}}=0$ (this doesn't exactly satisfy the mean value theorem, but I used this to disprove that $\lim_{x\to a}{\frac{f'(t)}{f'(x)}}=1$ in general). Are there any additional steps I should be taking? Or should I take an entirely different approach (if even possible without L'Hopital's rule)?

Answer 1

The function has to be twice differentiable at $\,x=a\,,\;$ otherwise there is a counterexample.

Let $\;f:\,]\!-\!\infty,+\infty[\to\Bbb R\;$ be the function defined as follows

$f(x)=\begin{cases} x^3\sin\left(\dfrac1x\right)\;,&\text{ if }\;x\neq0\\\\\;0\;\;,&\text{ if }\;x=0 \end{cases}$

It results that

$\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x^2}=\lim\limits_{x\to0}\,x\sin\left(\dfrac1x\right)=0$

but $\;f’’(0)\;$ does not exist, indeed

$f’(x)=\begin{cases} 3x^2\sin\left(\dfrac1x\right)-x\cos\left(\dfrac1x\right)\;,&\text{ if }\;x\neq0\\\\\;0\;\;,&\text{ if }\;x=0 \end{cases}$

and $\;\lim\limits_{x\to0}\dfrac{f’(x)-f’(0)}x= \lim\limits_{x\to0}\left[3x\sin\left(\dfrac1x\right)-\cos\left(\dfrac1x\right)\right]$

does not exist.

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Now, I will prove that $\,f’’(a)=0\,$ just assuming that there exists $\,f’’(a)\;$. It is not necessary that $\,f\,$ is twice differentiable at other points $\,x\neq a\,.$

Taylor’s Theorem with Peano’s form of remainder :
If $\,f\,$ is a function such that there exists its second derivative $\,f’’(a)\,$ at the point $\,x=a\,,\;$ then $$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}2f''(a)+R(x)$$ where $\,R(x)\,$ is a function such that $\,\lim\limits_{x\to a}\dfrac{R(x)}{(x-a)^2}=0\,.$

By applying Taylor’s Theorem with Peano’s form of remainder, we get that

$\dfrac{f(x)-f(a)}{(x-a)^2}=\dfrac{f’(a)}{x-a}+\dfrac{f’’(a)}2+\dfrac{R(x)}{(x-a)^2}\;.$

Since $\;\lim\limits_{x\to a}\dfrac{f(x)-f(a)}{(x-a)^2}=\lim\limits_{x\to a}\dfrac{R(x)}{(x-a)^2}=0\;,\;$ it follows that

$\lim\limits_{x\to a}\left[\dfrac{f’(a)}{x-a}+\dfrac{f’’(a)}2\right]=0\;,\;$ consequently ,

$f’(a)=f’’(a)=0\,.$

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Addendum :

It is possible to prove that $\,f’’(a)=0\,$ just applying the Mean Value Theorem in the following way.

$f’(a)=\lim\limits_{x\to a}\dfrac{f(x)\!-\!f(a)}{x\!-\!a}=\lim\limits_{x\to a}\left[(x\!-\!a)\dfrac{f(x)\!-\!f(a)}{(x\!-\!a)^2}\right]\!=\!0.$

Let $\,\delta\,$ be a real positive number ( $\delta\in\Bbb R^+$).
Let $\,g:[a-\delta,a+\delta]\to\Bbb R\,$ be the function defined as

$g(x)=\begin{cases} \dfrac{f(x)-f(a)}{x-a}\;\;,&\text{ if }\,x\in[a-\delta,a+\delta]\setminus\{a\}\\\\ f’(a)=0\;\;,&\text{ if }\,x=a \end{cases}$

$g’(a)=\lim\limits_{x\to a}\dfrac{g(x)-g(a)}{x-a}=\lim\limits_{x\to a}\dfrac{f(x)-f(a)}{(x-a)^2}=0\,.$

The function $\,g\,$ is continuous and differentiable on $\,[a-\delta,a+\delta]\,$ and it results that

$g’(x)=\begin{cases} \dfrac{f’(x)}{x-a}-\dfrac{f(x)-f(a)}{(x-a)^2}\;\;,&\text{ if }\,x\neq a\\\\ \;0\;\;,&\text{ if }\,x=a \end{cases}$

Consequently ,

$\dfrac{f’(x)}{x-a}=g’(x)+\dfrac{f(x)-f(a)}{(x-a)^2}\;$ $\;\forall x\!\in[a\!-\!\delta,a\!+\!\delta]\setminus\{a\}.$

For all $\,x\in[a-\delta,a+\delta]\setminus\{a\}\,,\,$ by applying the Mean Value Theorem to the function $\,g\,$ on the interval $\,I_x\,$ ( where $\,I_x=[a,x]\,$ if $\,x>a\,,\,$ otherwise $\,I_x=\,[x,a]\,),\,$ we get that there exists $\,\xi(x)\,$ in the interior of the interval $\,I_x\,$ such that

$\dfrac{g(x)-g(a)}{x-a}=g’\big(\xi(x)\big)\;.$

In this way, we get the function

$\;\xi(x):[a-\delta,a+\delta]\setminus\{a\}\to\;]a-\delta,a+\delta[\,\setminus\{a\}$

such that $\;\;\lim\limits_{x\to a}\,\xi(x)=a\;\;$ and

$g’\big(\xi(x)\big)=\dfrac{g(x)-g(a)}{x-a}\;\;\;\;\forall x\in[a-\delta,a+\delta]\setminus\{a\}\,.$

Moreover ,

$\begin{align} \dfrac{f’\big(\xi(x)\big)}{\xi(x)-a}&=g’\big(\xi(x)\big)+\dfrac{f\big(\xi(x)\big)-f(a)}{\big(\xi(x)-a\big)^2}=\\ &=\dfrac{g(x)-g(a)}{x-a}+\dfrac{f\big(\xi(x)\big)-f(a)}{\big(\xi(x)-a\big)^2}=\\ &=\dfrac{f(x)-f(a)}{(x-a)^2}+\dfrac{f\big(\xi(x)\big)-f(a)}{\big(\xi(x)-a\big)^2}\;\;,\quad\color{blue}{(*)}\\ &\forall x\!\in[a\!-\!\delta,a\!+\!\delta]\setminus\{a\}\,. \end{align}$

From the equality $\,(*)\,,\,$ it follows that

$\begin{align} f’’(a)&=\lim\limits_{x\to a}\dfrac{f’(x)\!-\!f’(a)}{x-a}=\lim\limits_{x\to a}\dfrac{f’(x)}{x\!-\!a}=\lim\limits_{x\to a}\dfrac{f’\!\big(\xi(x)\big)}{\xi(x)\!-\!a}=\\ &=\lim\limits_{x\to a}\left[\dfrac{f(x)-f(a)}{(x-a)^2}+\dfrac{f\big(\xi(x)\big)-f(a)}{\big(\xi(x)-a\big)^2}\right]=\\ &=0+0=0\;. \end{align}$

Answer 2

If $f$ is twice differentiable at $a,$ the conclusion follows by a single application of l'Hopital rule. First observe that $$f'(a)=\lim_{x\to a}{f(x)-f(a)\over x-a}=\lim_{x\to a}{f(x)-f(a)\over (x-a)^2}(x-a)=0$$ Next $$0=\lim_{x\to a}{f(x)-f(a)\over (x-a)^2}\overset{H}{=}{1\over 2}\lim_{x\to a}{f'(x) \over x-a}={1\over 2}\lim_{x\to a}{f'(x)-f'(a) \over x-a}={1\over 2}f''(a)$$ I do not see a simpler way. The proof of l'Hopital rule is based on the generalization of the mean value property to the ratios of the form $\displaystyle{f(x)-f(a)\over h(x)-h(a)}.$ In the OP case $h(x)=(x-a)^2.$

Remark There is a general theorem on Peano remainder applied in the OP case for $k=2.$ The proof of that theorem relies on l'Hopital rule applied $k-1$ times.

Answer 3

By translation of the variable and values, we are reduced to prove that if $f''(0)=\ell$ and $\lim_{x\to0}\frac{f(x)}{x^2}=0$ then $\ell=0.$

The assumptions imply the following Taylor developement of $f$ at $0$: $$f(x)=0+x\cdot0+\frac{x^2}2\ell+o(x^2)$$ and $\lim_{x\to0}\frac12\ell+o(1)=0,$ i.e. $\ell=0.$