# Two apparently different limits to the same question

I have a question with 2 possible solutions. and I can not decide which is right.

The question

Given $b>0$ and a function $f:(-b,b)\rightarrow\mathbb{R}$

, the second derivative exists at 0 and continuous at that point.

also known that $f'(0)\neq0$

Use taylor polynomial properties to calculate the following limit

$\underset{x\rightarrow0}{\lim}\frac{1}{f(x)-f(0)}-\frac{1}{xf'(0)}$

Using the taylor polynomial and the remainder we can say that

$f(x)=P_{1}(x)+R_{1}(x)$

i.e $f(x)=f(0)+x\cdot f'(x)+R_{1}(x)$

$\rightarrow R_{1}(x)=f(x)-f(0)-x\cdot f'(x)$

$\rightarrow-R_{1}(x)=f(0)+x\cdot f'(x)-f(x)$

Now distributing we get

$\frac{1}{f(x)-f(0)}-\frac{1}{xf'(0)}=\frac{x\cdot f'(x)+f(0)-f(x)}{xf'(0)\left(f(x)-f(0)\right)}=\frac{-R_{1}(x)}{xf'(0)\left(f(x)-f(0)\right)}$

*Here comes the doubt.

As a conclution from Cauchy's mean value theorem for functions we get that

$\underset{x\rightarrow0}{\lim}\frac{R_{1}(x)}{x}=0$ for this particular function.

Therefore : $\underset{x\rightarrow0}{\lim}\frac{1}{f(x)-f(0)}-\frac{1}{xf'(0)}=\underset{x\rightarrow0}{\lim}\frac{-R_{1}(x)}{xf'(0)\left(f(x)-f(0)\right)}=\overset{0}{\overbrace{\underset{x\rightarrow0}{\lim}\frac{R_{1}(x)}{x}}}\cdot\underset{x\rightarrow0}{\lim}\frac{1}{f'(0)\left(f(x)-f(0)\right)}$

Now to deal with the limit $\underset{x\rightarrow0}{\lim}\frac{1}{f'(0)\left(f(x)-f(0)\right)}$ I think to use the property

“if a fucntion has a derivative then it must be continuous at the point.”

Which means this function must have a limit at this point(in our case x=0)

If the function has a limit then it is bounded. i.e $\forall x\in(0-\delta,0+\delta)\,\,\,f(x)\leq M\in\mathbb{R}$

this means that the whole expression is bounded. $f'(0)\left(f(x)-f(0)\right)\leq f'(0)\left(M-f(0)\right)$

And consequently $\frac{1}{f'(0)\left(f(x)-f(0)\right)}$ is bounded...

Therefore a limit of a bounded function on an iterval and a function that goes to zero in the same interval is 0

(+)Second option using the lagrange remainder instead:

because the second derivative exists we can express the “l'agrange remainder” around zero as :

$R_{1}(x)=\frac{f''(c)}{2}x^{2} where \left|c-0\right|<\left|x-0\right|$

Therefore

$\frac{1}{f(x)-f(0)}-\frac{1}{xf'(0)}=\frac{-\frac{f''(c)}{2}x^{2}}{xf'(0)\left(f(x)-f(0)\right)}=\frac{-1}{2f'(0)}\cdot\frac{x}{\left(f(x)-f(0)\right)}\cdot f''(c)$

Calculating the limits

$\underset{x\rightarrow0}{\lim}\frac{-1}{2f'(0)}=\frac{-1}{2f'(0)}$

$\underset{x\rightarrow0}{\lim}\frac{x}{\left(f(x)-f(0)\right)}=\frac{x-0}{\left(f(x)-f(0)\right)}=\frac{1}{f'(0)}$

$\underset{x\rightarrow0}{\lim}f''(c)=f''(0)$ because$\left|c-0\right|<\left|x-0\right|$ and$f''$ is continuous.

$\underset{x\rightarrow0}{\lim}\frac{1}{f(x)-f(0)}-\frac{1}{xf'(0)}=\frac{-1}{2\left(f'(0)\right)^{2}}f''(0)$
• Consider $f(x)=(1+x)^2$ we have $\frac{1}{f(x)-f(0)}-\frac{1}{xf'(x)}=-\frac{1}{2(x+2)}$. The limit is $-1/4$ which is consistent with your second expression for the limit. – smcc Jul 30 '16 at 15:07
• "the second derivative exists at 0 and continuous at that point." Very strange hypothesis. Should read "the second derivative exists in a neighborhood 0 and is continuous at $0$." – zhw. Aug 29 '16 at 21:27
A much simpler approach is the direct evaluation of the limit. We have \begin{align} L &= \lim_{x \to 0}\frac{1}{f(x) - f(0)} - \frac{1}{xf'(0)}\notag\\ &= \lim_{x \to 0}\frac{xf'(0) - f(x) + f(0)}{xf'(0)\{f(x) - f(0)\}}\notag\\ &= -\lim_{x \to 0}\frac{f(x) - f(0) - xf'(0)}{x^{2}f'(0)}\cdot\frac{x}{f(x) - f(0)}\notag\\ &= -\frac{1}{\{f'(0)\}^{2}}\lim_{x \to 0}\frac{f(x) - f(0) - xf'(0)}{x^{2}}\notag\\ &= -\frac{f''(0)}{2\{f'(0)\}^{2}}\notag \end{align} The last step follows by Taylor's theorem with Peano's form of remainder which tells us that $$f(x) = f(0) + xf'(0) + \frac{f''(0)}{2}x^{2} + o(x^{2})$$ provided that $f''(0)$ exists. Therefore $$\lim_{x \to 0}\frac{f(x) - f(0) - xf'(0)}{x^{2}} = \frac{f''(0)}{2}$$ Instead of Taylor's theorem we can also apply L'Hospital's rule (only once) here and get the same answer.
Your first approach has serious flaws and the argument given there is incorrect (note that you have not used the existence of second derivative in your first approach). The expression $1/f'(0)\{f(x) - f(0)\}$ is in fact unbounded. You use the flawed idea that if $A$ is bounded above then $1/A$ is also bounded above.
Second approach is fine but it uses more hypotheses than required to solve the problem. You only need that $f'(0), f''(0)$ exist and $f'(0) \neq 0$ as I have shown in my answer.