Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$

Answer

$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$ [Dividing numerator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$
Applying the formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator respectively
Here, $\text{n}=\frac{2}{3},\text{m}=\frac{3}{4}$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{x}}}\frac{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}{\text{x}-\text{a}}}=\frac{\frac{2}{3}(\text{a})^{\frac{2}{3}-1}}{\frac{3}{4}(\text{a})^{\frac{3}{4}-1}}$
$=\frac{8}{9}\text{a}^{\frac{-1}{3}+\frac{1}{4}}$
$=\frac89\text{a}^{\frac{-1}{12}}$

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