Finding limit of $\frac{\tan{x}-\sin{x}}{x^3}$. I recently got this question in my yearly mathematics exam and was one of the questions I calculated wrong.
What I did was :-
$$\lim_{x\to 0}{\frac{\tan{x}-\sin{x}}{x^3}}$$
$$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$
$$=\lim_{x\to 0}{\frac{\to 0}{x^2}}-{\frac{\to 0}{x^2}}$$
$$=\lim_{x\to 0}{(\to 0)({\frac{1}{x^2}}-{\frac{1}{x^2}})}=0$$
which is wrong. The correct answer is $\frac{1}{2}$ as given by wolfram alpha.
So my question is :-
1) What's the error in my answer?
2) How is the correct answer to be calculated?
Thanks for help :)
 A: The following step in you calculation was wrong:
$$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$
$$=\lim_{x\to 0}{\frac{\to 0}{x^2}}-{\frac{\to 0}{x^2}}$$
You didn't write it down, but in order to do this you need to do an additional step, that is:
$$\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$
$$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-\lim_{x\to 0}{\frac{\sin{x}}{x^3}}.$$
But the above is not true. Separating sums in limits is only valid, if the limits of both summands exist and are finite, but in this case they don't.

A valid calculation of the limit would for example be:
\begin{align}
&\lim_{x\to 0}{\frac{\tan{x}-\sin x}{x^3}} \\
=& \lim_{x\to 0} \frac {\tan x}x \frac{1-\cos x}{x^2} \\
=& \lim_{x\to 0} \frac {\tan x}x \lim_{x\to 0}\frac{1-\cos x}{x^2} \\=&1\cdot \frac 1 2 =\frac 1 2.
\end{align}
Here we can calculate each limit separately and multiply them afterwards, as both exist.
(In the first step I just took the $\tan x$ factor outside).
A: Hint: Use the Taylor development: $sin(x)=x-x^3/6+O(x^3)$, $tan(x)=x+x^3/3+O(x^3)$.
A: we have $$\frac{-\sin(x){\cos(x)}+\sin(x)}{\cos(x)x^3}=\frac{\sin(x)(1-\cos^2(x))}{x^3\cos(x)(1+\cos(x))}=\frac{\sin^3(x)}{x^3}\frac{1}{\cos(x)(1+\cos(x))}$$
