Question about limits, derivative definition

Question

I have MacLaurin series: $$f(x)=f(0)+f'(0) x+\frac12 f''(0)x^2+...$$ Suppose $$f(x)=c_0+c_1 x+c_2 x^2+o(x^3)$$as $x\to0$. Then I can show $c_0=f(0)$ and $c_1=f'(0)$. Then $$c_2=\lim_{x\to0}(\frac{f(x)-f(0)-f'(0)x}{x^2})\\=\lim_{x\to0}(\frac{f'(x) }{x}-\frac{f'(0)}x)\\=f''(0)$$So the fraction of $\frac12$ has disappeared. Why? Was it that step when I introduced $f'(x)$ inside the limit? How can I make this correct?

Thanks for any help.

Answer 1

Because you made a mistake. You applied L'Hopital's rule, right?! But then use the fact that the derivative of $x^2$ is $2x$.

Answer 2

Your first equation:

$$c_2=\lim_{x\to0}\left(\frac{f(x)-f(0)-f'(0)x}{x^2}\right)$$

is correct. Then, you applied L'Hopital's rule to get:

$$c_2=\lim_{x\to0}\left(\frac{f'(x)-f'(0)}{2x}\right)$$

this is the correct second step.

Answer 3

$$\lim_{x \to 0} \frac{f(x)-f(0)}{x^2} \color{red}{ \neq} f''(0)$$

Instead, correct expression is $$\lim_{x \to 0} \frac{f(x)-f(0)}{x^2} = \frac{ f''(0)}{2}$$

You can see this by simply applying L'Hopitals Rule two times.