Question

Evaluate the following limit, L= $\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^{\dfrac{1}{2}}} - {2^{\dfrac{1}{2}}}}}{{{x^{\dfrac{1}{3}}} - {2^{\dfrac{1}{3}}}}}} \right)$.

Answer 1

Hint: First we have to use the first formula of the limit which is, $\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{{x^n} - {a^n}}}{{x - a}}} \right) = n{a^{n - 1}}$
Later we have to use laws of indices for the final calculation.

Complete step-by-step answer:
We have given, L = $\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^{\dfrac{1}{2}}} - {2^{\dfrac{1}{2}}}}}{{{x^{\dfrac{1}{3}}} - {2^{\dfrac{1}{3}}}}}} \right)$
Now, First multiply and divide the numerator and denominator by, $x - 2$
So we get, L = $\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\dfrac{{{x^{\dfrac{1}{2}}} - {2^{\dfrac{1}{2}}}}}{{x - 2}}}}{{\dfrac{{{x^{\dfrac{1}{3}}} - {2^{\dfrac{1}{3}}}}}{{x - 2}}}}} \right)$
Separate the limit to the both numerator and the denominator, we get
L = $\dfrac{{\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^{\dfrac{1}{2}}} - {2^{\dfrac{1}{2}}}}}{{x - 2}}} \right)}}{{\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^{\dfrac{1}{3}}} - {2^{\dfrac{1}{3}}}}}{{x - 2}}} \right)}}$
Now apply the formula $\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{{x^n} - {a^n}}}{{x - a}}} \right) = n{a^{n - 1}}$ to the numerator and denominator both,
So we get, L = $\dfrac{{\dfrac{1}{2}\left( {{2^{\dfrac{1}{2} - 1}}} \right)}}{{\dfrac{1}{3}\left( {{2^{\dfrac{1}{3} - 1}}} \right)}}$
Further solving we get,
L = $\dfrac{{3\left( {{2^{ - \dfrac{1}{2}}}} \right)}}{{2\left( {{2^{ - \dfrac{2}{3}}}} \right)}}$ (from $\dfrac{1}{2}$the 2 multiplies in to the denominator and in $\dfrac{1}{3}$ the 3 goes into the numerator)
By applying laws of the indices we get,
L = $\dfrac{3}{2}\left( {{2^{ - \dfrac{1}{2} + \dfrac{2}{3}}}} \right)$
L = $\dfrac{3}{2}\left( {{2^{\dfrac{1}{6}}}} \right)$

Therefore our final value of the given limit is L = $\dfrac{3}{2}\left( {{2^{\dfrac{1}{6}}}} \right)$

Note: We have to always separate limit function to the numerator and denominator both otherwise we cannot apply the formula.
When using laws of indices we have to change the proper sign of power otherwise we get the wrong answer.