Evaluate $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt{x}-\sqrt{2}}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}}$

In this limit problem, two algebraic expressions in irrational form formed a rational function.

$\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt{x}-\sqrt{2}}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}}$

In the numerator, the square root of $2$ is subtracted from square root of $x$. the cube root of $2$ is subtracted from the cube root of $x$ in the denominator. The limit of the rational function, which consists of irrational functions has to be evaluated as the value of $x$ is closer to $2$.

Evaluate the Limit by direct substitution

Use the direct substitution method to find the limit of the function as the value of $x$ tends to $2$.

$=\,\,\,$ $\dfrac{\sqrt{2}-\sqrt{2}}{\sqrt[\Large 3]{2}-\sqrt[\Large 3]{2}}$

$=\,\,\,$ $\dfrac{0}{0}$

The values of both numerator and denominator are zero as the value of $x$ approaches to $2$. Hence, the limit of the given function is indeterminate. So, the direct substitution method is not useful in this case. Therefore, let us try to calculate the limit in an alternative method.

Prepare the given function into required form

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}}$ $\times$ $1\bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}}$ $\times$ $\dfrac{x-2}{x-2}\bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{\sqrt{x}-\sqrt{2}}{x-2}}$ $\times$ $\dfrac{x-2}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}\bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt{x}-\sqrt{2}}{x-2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x-2}{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt{x}-\sqrt{2}}{x-2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{1}{\dfrac{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}{x-2}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt{x}-\sqrt{2}}{x-2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\sqrt[\Large 3]{x}-\sqrt[\Large 3]{2}}{x-2}}}$

D

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^{\Large \frac{1}{2}}-2^{\Large \frac{1}{2}}}{x-2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^{\Large \frac{1}{3}}-2^{\Large \frac{1}{3}}}{x-2}}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{1}{2}-1}$ $\times$ $\dfrac{1}{\dfrac{1}{3} \times (2)^{\displaystyle \dfrac{1}{3}-1}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{1-2 \times 1}{2}}$ $\times$ $\dfrac{1}{\dfrac{1}{3} \times (2)^{\displaystyle \dfrac{1-3 \times 1}{3}}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{1-2}{2}}$ $\times$ $\dfrac{1}{\dfrac{1}{3} \times (2)^{\displaystyle \dfrac{1-3}{3}}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $\dfrac{1}{\dfrac{1}{3} \times (2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $\dfrac{1}{\dfrac{1}{3}}$ $\times$ $\dfrac{1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{1}{2} \times (2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $3$ $\times$ $\dfrac{1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{1}{2}$ $\times$ $3$ $\times$ $(2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $\dfrac{1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{1 \times 3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $\dfrac{1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{-1}{2}}$ $\times$ $\dfrac{1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $\dfrac{(2)^{\displaystyle \dfrac{-1}{2}} \times 1}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $\dfrac{(2)^{\displaystyle \dfrac{-1}{2}}}{(2)^{\displaystyle \dfrac{-2}{3}}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{-1}{2}-\Big(\dfrac{-2}{3}\Big)}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle -\dfrac{1}{2}-\Big(-\dfrac{2}{3}\Big)}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle -\dfrac{1}{2}+\dfrac{2}{3}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{2}{3}-\dfrac{1}{2}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{2 \times 2-3 \times 1}{6}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{4-3}{6}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $(2)^{\displaystyle \dfrac{1}{6}}$

$=\,\,\,$ $\dfrac{3}{2}$ $\times$ $\sqrt[\Large 6]{2}$

$=\,\,\,$ $\dfrac{3}{2}$ $\sqrt[\Large 6]{2}$

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