lim as x - 0 of (sin x -x)/(tan x - x) using L' Hopitals Rule

Question

lim as x -> 0 of (sin x -x)/(tan x - x) using L' Hopitals Rule

anonymous · Answer

take derivative of the numerator and denominator independently.

anonymous · Answer

Somehow when I do that I keep getting 0/0 even after taking several derivatives.

goformit100 · Answer

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psymon · Answer

A bit tricky on this one, but when you turn everything into sines and cosines and do l'hopital's once, you can rearrange the equation in a way that it follows these two limit identities:

$$\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1  $$
$$\lim_{x \rightarrow 0}\frac{ cosx-1 }{ x }=0$$

anonymous · Answer

I'm still little confused how you can rearrange it

psymon · Answer

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May not be necessary to rearrange it like I did first, but it let me see things better like this. Either way, do l'hopital's rule from here then see if you can move things around to fit the identities above. Should work out.