Question

$limi{t}_{x\to a}\frac{{\left(a+2x\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3x\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(3a+x\right)}^{\frac{1}{3}}-{\left(4x\right)}^{\frac{1}{3}}}=$

• A. ${\left(\frac{2}{9}\right)}^{\frac{4}{3}}$
• B. ${\left(\frac{2}{3}\right)}^{\frac{4}{3}}$
• C. $\left(\frac{2}{3}\right){\left(\frac{2}{9}\right)}^{\frac{4}{3}}$
• D. $\left(\frac{2}{9}\right){\left(\frac{2}{3}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is C

$\left(\frac{2}{3}\right){\left(\frac{2}{9}\right)}^{\frac{4}{3}}$

Explanation for Correct Answer:

Evaluating the given function:

$$\begin{gathered} limi{t}_{x\to a}\frac{{\left(a+2x\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3x\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(3a+x\right)}^{\frac{1}{3}}-{\left(4x\right)}^{\frac{1}{3}}}=limi{t}_{x\to a}\frac{{\left(a+2a\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3a\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(3a+a\right)}^{\frac{1}{3}}-{\left(4a\right)}^{\frac{1}{3}}} \\ =limi{t}_{x\to a}\frac{{\left(3a\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3a\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(4a\right)}^{\frac{1}{3}}-{\left(4a\right)}^{\frac{1}{3}}} \\ =\frac{0}{0} \end{gathered}$$

$\frac{0}{0}$ indeterminant form using L Hospitals rule.

$$\begin{gathered} limi{t}_{x\to a}\frac{{\left(a+2x\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3x\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(3a+x\right)}^{\frac{1}{3}}-{\left(4x\right)}^{\frac{1}{3}}}=limi{t}_{x\to a}\frac{\left(\frac{1}{3}{\left(a+2x\right)}^{{\displaystyle \frac{-2}{3}}}.2\right)-\left(\frac{1}{3}{\left(3x\right)}^{\frac{-2}{3}}.3\right)}{\left(\frac{1}{3}{\left(3a+x\right)}^{\frac{-2}{3}}\right)-\left(\frac{1}{3}{\left(4x\right)}^{\frac{-2}{3}}.4\right)}\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right] \\ =limi{t}_{x\to a}\frac{\cancel{\frac{1}{3}}\left(2.{\left(a+2x\right)}^{{\displaystyle \frac{-2}{3}}}-3.{\left(3x\right)}^{\frac{-2}{3}}\right)}{\cancel{\frac{1}{3}}\left({\left(3a+x\right)}^{\frac{-2}{3}}-4.{\left(4x\right)}^{\frac{-2}{3}}\right)} \end{gathered}$$

$$\begin{gathered} =\frac{2.{\left(a+2a\right)}^{{\displaystyle \frac{-2}{3}}}-3.{\left(3a\right)}^{\frac{-2}{3}}}{{\left(3a+a\right)}^{\frac{-2}{3}}-4.{\left(4a\right)}^{\frac{-2}{3}}} \\ =\frac{2.{\left(3a\right)}^{{\displaystyle \frac{-2}{3}}}-3.{\left(3a\right)}^{\frac{-2}{3}}}{{\left(4a\right)}^{\frac{-2}{3}}-4.{\left(4a\right)}^{\frac{-2}{3}}} \\ =\frac{-1.{\left(3\right)}^{\frac{-2}{3}}\cancel{{\left(a\right)}^{\frac{-2}{3}}}}{-3.{\left(4\right)}^{\frac{-2}{3}}\cancel{{\left(a\right)}^{\frac{-2}{3}}}} \end{gathered}$$

$$\begin{gathered} =\frac{1.{\left(3\right)}^{\frac{-2}{3}}}{3.{\left(2^2\right)}^{\frac{-2}{3}}} \\ =\frac{1}{3}.{\left(\frac{3}{4}\right)}^{\frac{-2}{3}} \\ =\frac{1}{3}.{\left({\left(\frac{4}{3}\right)}^2\right)}^{\frac{1}{3}}\left[\because a^{-1}=\frac{1}{a}\right] \\ =\frac{1}{3}.{\left(\frac{16}{9}\right)}^{\frac{1}{3}} \\ =\frac{1}{3}.{\left(\frac{8\times 2}{9}\right)}^{\frac{1}{3}} \\ =\frac{2}{3}.{\left(\frac{2}{9}\right)}^{\frac{1}{3}}\left[\because 8^{\frac{1}{3}}=2\right] \end{gathered}$$

Hence, the correct answer is option (C).