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**Hint:**Here, we have to find the value of \[\mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}\]. If we take the limit $x \to 1$ we get $\dfrac{0}{0}$ which is an indeterminate form, so its limit can be calculated by the “L’HOPITAL’S” rule. Firstly, differentiate the numerator and denominator with respect to $x$ then find the limit $x \to 1$.

**Complete step-by-step solution:**

Given, we have to find the value of \[\mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}\].

By taking the limit $x \to 1$ of $\dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}$ we get $\dfrac{0}{0}$ which is an indeterminate form.

Here, we have to apply the “L’HOPITAL’S” rule. According to which we have to differentiate the numerator and differentiate the denominator with respect to $x$. So, by differentiating we can write \[\mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}\] as

$

\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \left( { - \dfrac{1}{3}{x^{ - \dfrac{1}{3} - 1}}} \right)}}{{ - \left( { - \dfrac{2}{3}{x^{ - \dfrac{2}{3} - 1}}} \right)}} \\

\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}}}}{{\dfrac{2}{3}{x^{\dfrac{{ - 5}}{3}}}}}

$

Now, by taking limit $x \to 1$ we get

$ \Rightarrow \dfrac{{\dfrac{1}{3}\left( 1 \right)}}{{\dfrac{2}{3}\left( 1 \right)}} = \dfrac{1}{2}$

**Thus, the value of \[\mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}\] is $\dfrac{1}{2}$.**

**Note:**L’HOPITL’S Rule:

If $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$ or $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{\infty }{\infty }$, then $\mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ is the required result.

So, this rule tells us that if we get $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ which is an indeterminate form then all we need to do is differentiate the numerator and differentiate the denominator and then take the limit $x \to a$which gives the required result.

Alternate method:

This can be solved by simply arranging the terms of the numerator and the denominator. We observed that the denominator of the given fraction is $1 - {x^{\dfrac{{ - 2}}{3}}}$ which can also be written as ${\left( 1 \right)^2} - {\left( {{x^{\dfrac{{ - 1}}{3}}}} \right)^2}$. Then, by using a mathematical identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we can write ${\left( 1 \right)^2} - {\left( {{x^{\dfrac{{ - 1}}{3}}}} \right)^2}$as $\left( {1 - {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {1 + {x^{\dfrac{{ - 1}}{3}}}} \right)$.

So, the given fraction $\dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}$ can be written as $\dfrac{{\left( {1 - {x^{\dfrac{{ - 1}}{3}}}} \right)}}{{\left( {1 - {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {1 + {x^{\dfrac{{ - 1}}{3}}}} \right)}}$.

The term $\left( {1 - {x^{\dfrac{{ - 1}}{3}}}} \right)$ is present in both the numerator and the denominator so, it is cancelled out and the \[\mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^{\dfrac{{ - 1}}{3}}}}}{{1 - {x^{\dfrac{{ - 2}}{3}}}}}\] can be written as $\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{1 + {x^{\dfrac{{ - 1}}{3}}}}}$. Then we have to take its limit $x \to 1$.

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