# How can I evaluate lim_(x->0) (sinx-x)/x^3 without using L'Hopital's rule?

Dec 23, 2017

$- \frac{1}{6}$.

#### Explanation:

Suppose that $\text{the Reqd. Limit L=} {\lim}_{x \to 0} \frac{\sin x - x}{x} ^ 3$.

Substiture $x = 3 y , \text{ so that, as } x \to 0 , y \to 0$.

$\therefore L = {\lim}_{y \to 0} \frac{\sin 3 y - 3 y}{{\left(3 y\right)}^{3}}$,

$= {\lim}_{y \to 0} \frac{\left(3 \sin y - 4 {\sin}^{3} y\right) - 3 y}{27 {y}^{3}}$,

$= {\lim}_{y \to 0} \left\{\frac{3 \left(\sin y - y\right)}{27 {y}^{3}} - \frac{4 {\sin}^{3} y}{27 {y}^{3}}\right\}$,

$\Rightarrow L = {\lim}_{y \to 0} \frac{1}{9} \cdot \left(\frac{\sin y - y}{y} ^ 3\right) - \frac{4}{27} \cdot {\left(\sin \frac{y}{y}\right)}^{3.} . . \left(\ast\right)$.

Note that, here,

${\lim}_{y \to 0} \left(\frac{\sin y - y}{y} ^ 3\right) = {\lim}_{x \to 0} \left(\frac{\sin x - x}{x} ^ 3\right) = L$.

Therefore, $\left(\ast\right) \Rightarrow L = \frac{1}{9} \cdot L - \frac{4}{27} , \mathmr{and} , \frac{8}{9} L = - \frac{4}{27}$.

Hence, $L = - \frac{4}{27} \cdot \frac{9}{8} = - \frac{1}{6}$.

Enjoy Maths.!

Dec 24, 2017

Kindly refer to the Explanation.

#### Explanation:

In this Second Method, we use the following series :

sinx=x-x^3/3!+x^5/5!-...oo.

 :. sinx -x=-x^3/3!+x^5/5!-x^7/7!+...oo.

:. (sinx-x)/x^3=-1/6+x^2/5!-x^5/7!+...oo.

$\Rightarrow {\lim}_{x \to 0} \frac{\sin x - x}{x} ^ 3 = - \frac{1}{6} + 0 - 0 + \ldots \infty$

$i . e . , {\lim}_{x \to 0} \frac{\sin x - x}{x} ^ 3 = - \frac{1}{6} ,$ as in Method 1.