Question

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$.

Answer 1

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}=\frac{1}{2}$
Transform the function in this way:
$\frac{\tan x-\sin x}{x^3}=\frac{1}{x^3}(\frac{\sin x}{\cos x}-\sin x)$
$\frac{\tan x-\sin x}{x^3}=\frac{1}{x^3}\left(\frac{\sin x-\sin x\cos x}{\cos x}\right)$
$\frac{\tan x-\sin x}{x^3}=\frac{\sin x}{x^3}\frac{1-\cos x}{\cos x}$
$\frac{\tan x-\sin x}{x^3}=\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^2}\right)\left(\frac{1}{\cos x}\right)$
We can use now the well known trigonometric limit:
$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$
and using the trigonometric identity:
${\sin }^2\alpha =\frac{1-\cos 2\alpha }{2}$
we have,
$\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=\underset{x\to 0}{lim}\frac{2{\sin }^2\left(\frac{x}{2}\right)}{x^2}=\frac{1}{2}\underset{x\to 0}{lim}{\left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)}^2=\frac{1}{2}$
While the third function is continuous so:
$\underset{x\to 0}{lim}\frac{1}{\cos x}=\frac{1}{1}=1$
and we can conclude that:
$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}=\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^2}\right)\left(\frac{1}{\cos x}\right)=1\times \frac{1}{2}\times 1=\frac{1}{2}$
graph $\frac{\tan x-\sin x}{x^3}[-1.25,1.25,-0.025,1]$