Answer $A$
Given:
$$\underset{x \rightarrow 1}{\lim} \dfrac{x^{\frac{1}{3}} -1}{x^{\frac{1}{4}}-1}\qquad \rightarrow(1)$$
On applying L'Hospital in $(1)$, we get:
$$ =\underset{x \rightarrow 1}{\lim} \dfrac{\frac{dy}{dx}(x^{\frac{1}{3}} -1)}{\frac{dy}{dx}(x^{\frac{1}{4}}-1)} $$
$$ =\underset{x \rightarrow 1}{\lim} \dfrac{\frac{1}{3}x^{{\frac{1}{3}-1}} -1}{\frac{1}{4}x^{{\frac{1}{4}}-1}-1} $$
$$=\underset{x \rightarrow 1}{\lim} \dfrac{\frac{1}{3}x^{{\frac{-2}{3}}}}{\frac{1}{4}x^{{\frac{-3}{4}}}} $$
Now, on substituting $x = 1$ above, we get:
$$ = \dfrac{\frac{1}{3}}{\frac{1}{4}}$$
$$ = \dfrac{4}{3}$$
$\therefore\; A$ is the correct option.