# Without using L'Hospital's rule, what is $\lim\limits_{x \to 0}\dfrac{\tan{x}-\sin{x}}{x^3}$?

I have to find the following limit without using L'Hôpital's rule:

$$\lim\limits_{x \to 0}\dfrac{\tan{x}-\sin{x}}{x^3}$$

I first converted $$\tan{x}$$ to $$\frac{\sin{x}}{\cos{x}}$$, and factored like this:

$$\dfrac{\dfrac{\sin{x}}{\cos{x}}-\sin{x}}{x^3}$$ $$\Rightarrow \dfrac{1}{x}\dfrac{\sin{x}}{x}\dfrac{1}{\cos{x}}\left(\dfrac{1-\cos{x}}{x}\right)$$

I can find the limit of each fraction except the first which is $$\frac{1}{x}$$.

How can I solve this without using L'Hospital's rule?

• Hint. $\frac{\sin{x}}{x}\frac{1}{\cos{x}}\frac{1-\cos^2(x)}{x^2}\frac{1}{1+\cos(x)}=\left(\frac{\sin{x}}{x}\right)^3\frac{1}{\cos{x}}\frac{1}{1+\cos(x)}$ Apr 27, 2021 at 11:50

I would do it this way: $$\frac{\tan{x}-\sin{x}}{x^3}=\dfrac{\sin x(1-\cos x)}{x^3\cos x}=\dfrac{\sin x(1-\cos^2x)}{x^3\cos x(1+\cos x)}=\frac{\sin^3x}{x^3}\,\frac 1{\cos x(1+\cos x)}.$$
Added: it may be shortened, using the result of this standard high-school exercise: $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12.$$
$$1-\cos x=2\sin^{2}(x/2)$$, so you need $$x^{2}$$ in the denominator to have the pairing.
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to0}\frac{\tan x}x\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12\lim_{x\to0}\frac{\sin^2\frac x2}{\left(\frac x2\right)^2}.$$
We note that as $$x \to 0$$ $$\frac{\tan x -\sin x}{x^3}=\frac{\tan x(1-\cos x)}{x^3} \sim \frac{\tan x \,(\frac{x^2}{2})}{x^3} \sim \frac{1}{2}$$