How can I solve this limit? lim x---->0 (sin^3x) /(x-sinx)^3

1 Answer
Jun 28, 2018

${\lim}_{x \to 0} \frac{{\sin}^{3} x}{x - \sin x} ^ 3 = + \infty$

Explanation:

Divide and multiply by ${x}^{3}$:

$\frac{{\sin}^{3} x}{x - \sin x} ^ 3 = {\sin}^{3} \frac{x}{x} ^ 3 \cdot {x}^{3} / {\left(x - \sin x\right)}^{3}$

$\frac{{\sin}^{3} x}{x - \sin x} ^ 3 = {\left(\sin \frac{x}{x}\right)}^{3} \cdot {\left(\frac{1}{\frac{x - \sin x}{x}}\right)}^{3}$

$\frac{{\sin}^{3} x}{x - \sin x} ^ 3 = {\left(\sin \frac{x}{x}\right)}^{3} \cdot {\left(\frac{1}{1 - \sin \frac{x}{x}}\right)}^{3}$

Using the well known trigonometric limit:

${\lim}_{x \to 0} \sin \frac{x}{x} = 1$

and then:

${\lim}_{x \to 0} {\left(\sin \frac{x}{x}\right)}^{3} = 1$

On the other hand to evaluate:

${\lim}_{x \to 0} {\left(\frac{1}{1 - \sin \frac{x}{x}}\right)}^{3}$

note that the numerator is finite and the denominator tends to zero and is always positive because $\sin x < x$. Then:

${\lim}_{x \to 0} {\left(\frac{1}{1 - \sin \frac{x}{x}}\right)}^{3} = + \infty$

and so:

${\lim}_{x \to 0} \frac{{\sin}^{3} x}{x - \sin x} ^ 3 = + \infty$