Question

Identify the value of $\underset{x\to 1}{lim}\frac{1-x^{1/3}}{1-x^{1/4}}$

• A. $1$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $0$

Answer 1

Answer: (B)

$\frac{4}{3}$

Let $L={lim}_{x\to 1}\frac{1-x^{1/3}}{1-x^{1/4}}$

This is $\frac{0}{0}$ form, therefore by using L'Hospital rule

Therefore, $L={lim}_{x\to 1}\frac{0-{\frac{1}{3}x}^{-2/3}}{0-\frac{1}{4}{x}^{-3/4}}$

$\Rightarrow L={lim}_{x\to 1}\frac{4}{3}\frac{x^{-2/3}}{x^{-3/4}}\Rightarrow L=\frac{4}{3}\frac{{\left(1\right)}^{-2/3}}{{\left(1\right)}^{-3/4}}\Rightarrow L=\frac{4}{3}$

.Option B is correct.