# lim_(trarr1){t-1}/(t^3-1)= ?

Jul 21, 2016

$\frac{1}{3}$

#### Explanation:

Making $x = {z}^{3}$

${\lim}_{x \rightarrow 1} \frac{\sqrt[3]{x} - 1}{x - 1} \equiv {\lim}_{z \to 1} \frac{z - 1}{{z}^{3} - 1}$

but

$\frac{z - 1}{{z}^{3} - 1} = \frac{1}{{z}^{2} + z + 1}$

so

${\lim}_{x \rightarrow 1} \frac{\sqrt[3]{x} - 1}{x - 1} = \frac{1}{3}$

Jul 21, 2016

$\frac{1}{3}$.

#### Explanation:

In fact, it is a Standard Form of Limit that

${\lim}_{x \rightarrow a} \frac{{x}^{n} - {a}^{n}}{x - a} = n \cdot {a}^{n - 1} , w h e r e , n \in \mathbb{R}$.

In our example, $a = 1 , n = \frac{1}{3}$.

The Reqd. Limit$= \frac{1}{3} \cdot {\left(1\right)}^{\frac{1}{3} - 1} = \frac{1}{3}$.

Aliter :-

Take substn. $x = {t}^{3}$, so that, as $x \rightarrow 1 , t \rightarrow 1$.

$\therefore$ Reqd. Limit$= {\lim}_{t \rightarrow 1} \frac{\sqrt[3]{{t}^{3}} - 1}{{t}^{3} - 1}$

$= {\lim}_{t \rightarrow 1} \frac{t - 1}{\left(t - 1\right) \left({t}^{2} + t + 1\right)}$

As $t \rightarrow 1 , t \ne 1$, so, $\left(t - 1\right) \ne 0$ can be cancelled, to get,

The Reqd. Lim.$= {\lim}_{t \rightarrow 1} \frac{1}{{t}^{2} + t + 1} = \frac{1}{3}$, as before!

Enjoy Maths.!