Finding the limit of: $\lim_{x\rightarrow 0} \left(\frac{1}{f(x)-f(0)}- \frac{1}{xf'(0)} \right)$ using taylor polynomials no solution provided so I was hoping someone would do a quick look over and make sure it looks ok.
Finding the limit of:
$$\lim_{x\rightarrow 0} \left(\frac{1}{f(x)-f(0)}- \frac{1}{xf'(0)} \right)$$
Using taylor polynomials, where $f:(-b,b)\rightarrow R$, a function that has a second derivative that is continuous at $x=0$, and $f'(0)\ne0$.  
We'll derive the taylor polynomial around x=0:
$$P_f(x)= f(0)+f'(0)x+ \frac{f^{2}(x^2)}{2}$$
Let's go back to our limit, and with some algebraic manipulation we get:
$$\frac{xf'(0)-f(x)+f(0)}{(f(x)-f(0))(xf'(0))}$$
We'll take thee terms from our taylor polynomial (two didn't work out! got 0/0) and plug them in:
$$\frac{xf'(0)-f(0)+f'(0)x+f(0)}{(f(0)+f'(0)x+-f(0))(xf'(0))}$$
A little cleaning up:
$$\frac{xf'(0)-f(0)-f'(0)x- \frac{f''(0)(x^2)}{2}+f(0)}{(f(0)+f'(0)x+ \frac{f''(0)(x^2)}{2}-f(0))(xf'(0))}$$
$$\frac{-\frac{f''(0)(x^2)}{2}} {(f'(0)x+ \frac{f''(0)(x^2)}{2})(f'(0)x)}$$
We'll open up the brackets in the denominator, and factor out an x^2:
$$\frac{-\frac{f''(0)}{2}} {((f'(0))(f'(0)))+ \frac{f''(0)((f'(0)(x))}{2}}$$
Now we'll take the limit:
$$\lim_{x\rightarrow 0}=\frac{-\frac{f''(0)}{2}} {((f'(0))(f'(0)))+ x\left[\frac{f''(0)((f'(0))}{2}\right]} =\frac{-f''(0)}{2((f'(0))(f'(0)))}$$
To whoever read this, thank you. I'm sure reading all that was worse than writing it. Does this seem right? 
 A: If we assume $f$ to be twice differentiable near $x=0$, we may use Taylor's Theorem with the Lagrange form of the remainder to write$$f(x)=f(0)+f'(0)\cdot x+\frac{f''\left(\xi_L\right)}{2}\cdot x^2$$for some $\xi_L$ between $0$ and $x$. Then the limit expression may be written as:
$$\frac{f'(0)\cdot x-f(x)+f(0)}{\big(f(x)-f(0)\big)\cdot \big(f'(0)\cdot x\big)}=\frac{-\frac{1}{2}\cdot f''\left(\xi_L\right)\cdot x^2}{\big(f'(0)\cdot x + \frac{1}{2}\cdot f''\left(\xi_L\right)\cdot x^2\big)\cdot\big(f'(0)\cdot x\big)}$$
Divide numerator and denominator by $x^2$ to obtain:
$$\frac{-\frac{1}{2}\cdot f''\left(\xi_L\right)}{\big(f'(0) + \frac{1}{2}\cdot f''\left(\xi_L\right)\cdot x\big)\cdot f'(0)}$$
So that, as $x\to 0$, so too does $\xi_L \to 0$; the continuity of $f''$ near $x=0$ then implies that the limit approaches $$\frac{-\frac{1}{2}\cdot f''\left(0\right)}{{\left(f'(0)\right)}^2}$$
which is always well-defined since we assumed $f'(0)\neq 0$. In short, your calculations are correct, but this is a more rigorous formalization of your hard-to-read solution.
A: Rewrite the function as
$$
\frac{xf'(0)-f(x)+f(0)}{xf'(0)(f(x)-f(0))}=
\frac{xf'(0)-f(0)-xf'(0)-\frac{x^2}{2}f''(0)+f(0)+o(x^2)}
  {xf'(0)(xf'(0)+o(x))}=
\frac{-\frac{x^2}{2}f''(0)+o(x^2)}{x^2f'(0)^2+o(x^2)}
$$
You're making some confusion in your steps, particularly when you forget the second order term and plug it in back (before “A little cleaning up”). The remainder term should not be omitted.
