Find $\lim\limits_{x\rightarrow 0} \frac{e^x -x -1}{1-\cos(x)}$ with taylor

Question

Find $$\lim_{x\rightarrow 0} \frac{e^x -x -1}{1-\cos(x)}$$ with taylor

My try and stuck

let $$ f(x) = e^x - x -1$$ and $$ g(x) = 1-\cos(x)$$

Now, I am taylor pattern with penao rest for $f$:
$$f'(x) = e^x - 1 \rightarrow f'(0)=0 \\ f''(x) = e^x \rightarrow f''(0)=1 $$ and futher there is the same situation so I stop there. Ok - now $g$: $$g'(x) = 1+\sin(x) \rightarrow g'(0)=1 $$ $$g''(x) = 1+\cos(x) \rightarrow g''(0)=2 $$ Ok, I stop there too because I want to have the same degree

$$ \frac{e^x -x -1}{1-\cos(x)} = \frac{0+0+\frac{1}{2}x^2+\frac{r(x)}{x^2}}{0+x+\frac{2}{2!}x^2 + p(x)} = \\ \frac{\frac{1}{2}+\frac{r(x)}{x^2}}{\frac{1}{x}+1+\frac{p(x)}{x^2}}$$

and there I stuck - what can be done with $\frac{1}{x}$ in the denominator?

Answer 1

should be $g'(x) = \sin(x) $ and $g''(x) = \cos(x)$

Answer 2

Arround zero, $$e^x=1+x+\frac{x^2}{2}(1+\epsilon(x))$$ and $$1-\cos(x)\sim \frac{x^2}{2}$$

thus, the limit becomes

$$\lim_{x\to 0}\frac{\frac{x^2}{2}(1+\epsilon(x))}{\frac{x^2}{2}}=1$$