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?
Best Answer
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: $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12.$$