This is what I did:Find the value of $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}$.

$\displaystyle f(x)=sin(x)-x$

$\displaystyle f'(x)=cos(x)-1$

$\displaystyle f^{(2)}(x)=-sin(x)$

$\displaystyle f^{(3)}(x)=-cos(x)$

$\displaystyle g(x)=tan(x)-x$

$\displaystyle g'(x)=sec^2(x)-1$

$\displaystyle g^{(2)}(x)=\frac{2sin(x)}{cos^3(x)}$

$\displaystyle g^{(3)}(x)=\frac{2cos^2(x)+6sin^2(x)}{cos^4(x)}$

Hence:

$\displaystyle a_3=f^{(3)}(0)=-cos(0)=-1$

$\displaystyle b_3=g^{(3)}(0)=2$

Hence $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}=\frac{a_3}{b_3}=-\frac{1}{2}$

However, when I put x=0.0000000001 into my calculator (as a check) I end up with 1.

What's going wrong??? :S