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$\underset{x\rightarrow 1}{\lim}\dfrac{x^{\frac{1}{3}}-1}{x^{\frac{1}{4}}-1}$ equals

  1. $\frac{4}{3}$
  2. $\frac{3}{4}$
  3. $1$
  4. None of these
in Calculus by Boss (17.5k points)
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1 Answer

+1 vote

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.

by Boss (19.1k points)
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