Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$.

Question

I have a difficulty in calculating this limit:

$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$

I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$

Then I got stucked, could anyone help me in solving it?

Answer 1

For $x\ne0,$

$${\tan x-\sin x\over x^3}=\left({\sin x\over x}\right)^3\dfrac1{\cos x \,(1+\cos x)}$$

Now as $x\to0,x\ne0$

Answer 2

$$\frac{\tan{x}-\sin{x}}{x^3}=\frac{x+\frac{1}{3}x^3+o(x^3) - (x-\frac{1}{6}x^3+o(x^3))}{x^3}=\frac{\frac{1}{2}x^3+o(x^3)}{x^3}=\frac{1}{2} +\frac{o(x^3)}{x^3}\stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{2}$$

Answer 3

HINT:

We can write

$$\begin{align} \frac{\tan(x)-\sin(x)}{x^3}&=\frac{\sin(x)(1-\cos(x))}{\cos(x)x^3}\\\\ &=\left(\frac{1}{\cos(x)}\right)\left(\frac{\sin(x)}{x}\right)\left(\frac{1-\cos(x)}{x^2}\right) \end{align}$$

Answer 4

If we let $A_n$ be the up/down numbers we have:

$$\tan x = \sum_{n=0}^\infty\frac{A_{2n+1}}{(2n+1)!}x^{2n+1}$$ $$\sin x = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Thus the first few coefficients of $\tan x - \sin x$ are:

$$0 + \frac{1}{2}x^3 + \frac{1}{8}x^5 + \cdots$$

Thus if we divide by $x^3$ we get a constant coefficient of $\dfrac{1}{2}$ and everything else vanishes as $x \to 0$.