Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
 A: HINT:
$\lim_{x\to0}\left(\left(\dfrac{\tan x}x\right)^3\cdot\dfrac1{x^2}-\left(\dfrac{\sin x}x\right)^3\cdot\dfrac1{x^2}\right)$ is of the form $\infty-\infty$
See List of indeterminate forms
Use 
$$\dfrac1{\cos^3x}\cdot\left(\dfrac{\sin x}x\right)^3\cdot\dfrac{1-\cos x}{x^2}\cdot(1+\cos x+\cos^2x)$$
A: $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}$$
with $\cos x\approx 1-\frac12x^2$ as $x\to0$
$$\lim_{x\to 0} \dfrac{\sin ^3 x}{x^3}\times\dfrac{1-(1-\frac12x^2)^3}{x^2\times1}=\lim_{x\to 0}\frac32-\frac34x^2+\frac18x^4=\frac32$$
A: $$\begin{align}\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}&=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \left(\dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\cdot\frac{1+\cos x}{1+\cos x}\right), \quad\text{note that }(1-\cos x)(1+\cos x)=\sin^2x\\
&=\lim_{x\to 0} \dfrac{\sin ^5 x\quad(1+\cos x+\cos^2x)}{x^5\cos ^3 x\quad(1+\cos x)}\\
&=\lim_{x\to 0}\left[\left(\frac{\sin x}{x}\right)^5\cdot \frac{1}{\cos^3x}\cdot\frac{1+\cos x+\cos^2x}{1+\cos x}\right]\\
&=1^5\cdot\frac{1}{1^3}\cdot\frac{1+1+1^2}{1+1}=\frac{3}{2}.
\end{align}$$
A: $$\dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\ldots=\frac{\sin^3x\left(1-\cos^3x\right)}{x^5\cos^3x}=\tan^3x \cdot\frac{1-\cos^3x}{x^5}$$ $$\sim x^3 \cdot\frac{1-\cos^3x}{x^5}=\frac{1-\cos^3x}{x^2}\sim \frac{1-\left(1-x^2/2+o(x^2)\right)^3}{x^2}$$ $$=\frac{3x^2/2+o(x^2)}{x^2}=\frac{3}{2}+\frac{o(x^2)}{x^2}\underbrace{\to}_{\text{ as }x\to 0}\frac{3}{2}+0=\frac{3}{2}$$
A: $$\frac{\tan ^3 x - \sin ^3 x}{x^5}=\tan^3x\dfrac{1 - \cos ^3 x}{x^5}=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2}\right)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{1-\cos x}{x^2}\right)(1+\cos x+\cos^2x)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{\sin x}{x}\right)^2\left(\dfrac{1}{1+\cos x}\right)(1+\cos x+\cos^2x)$$
Now use  the fundamental limits:
$$\lim_{x\to 0}\frac{\sin x }{x}=1=\lim_{x\to 0}\frac{\tan x }{x}$$
