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? 

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.
