Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$
I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\frac{\tan x}{x}-\frac{1}{x^2}}\cdot\lim{\frac{\sin x}{x}}\\
=\lim{\frac{1}{x^2}}-\lim{\frac{1}{x^2}}
=0$$
The answer using the standard method gives $\frac{1}{2}$.
It would be great if someone could explain to me why this method is wrong.
1) With l'Hospital:
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac1{\cos^2x}-\cos x}{3x^2}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac{2\sin x}{\cos^3x}+\sin x}{6x}=$$
$$=\lim_{x\to0}\frac16\frac{\sin x}x\left(\frac2{\cos^3x}+1\right)=\frac16\cdot1\cdot(2+1)=\frac12$$
2) Without l'Hospital:
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}x\frac{1-\cos x}{x^2}\frac1{\cos x}=1\cdot\frac12\cdot1=\frac12$$
Where we used
$$\frac{1-\cos x}{x^2}=\frac{\sin^2x}{(1+\cos x)x^2}=\frac1{1+\cos x}\left(\frac {\sin x}x\right)^2\xrightarrow[x\to0]{}\frac12\cdot1^2=\frac12$$
$\frac{\tan{x}-\sin{x}}{x^3}=\frac{2\sin{x}\sin^2\frac{x}{2}}{x^3\cos{x}}\rightarrow2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$
