How to get the limit (without l'hopital) of
(sin (x) - x)/(tan(x)-x) limit x approaches 0.

Question

How to get the limit (without l'hopital) of
(sin (x) - x)/(tan(x)-x) limit x approaches 0.

anonymous · Answer

$$\sin (x)=x-\frac{x^3}{6}+O\left(x^5\right)\
\tan (x)=x+\frac{x^3}{3}+O\left(x^5\right)$$

anonymous · Answer

uh oh... i forget to try this with Taylor ,but i was hoping for something using only substitution or cancelling.
any way thanks a lot man.

anonymous · Answer

The limit is $ - \frac 1 2$

anonymous · Answer

Ah, my mistake!

anonymous · Answer

Because
$$
\frac { \frac{-x^3}6}

{\frac{x^3}{3}}=-\frac 1 2
$$

anonymous · Answer

Actually, using Series is like using derivatives and derivatives are used in L'Hospital's Rule

anonymous · Answer

Thus this is the only way to get it.
Can't anyone find a simple way??

xapproachesinfinity · Answer

you can separate the fraction an manipulate it little bit
to get the usual limit x/sinx
tanx/x and 1/cosx and so on i guess the result is 1-1/2=-1/2

ganeshie8 · Answer

divide below two limits : http://math.stackexchange.com/questions/217081/determine-lim-x-to-0-fracx-sinxx3-frac16-without-lhospita?rq=1 http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3?lq=1

anonymous · Answer

Thanks a lot  man.