If the solve the problem
Question:

$\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0)$ is equal to :

  1. $\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}$

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

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

  4. $\left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}}$


Correct Option: 1

Solution:

Required limit

$L=\lim _{h \rightarrow 0} \frac{(a+2(a+h))^{1 / 3}-(3(a+h))^{1 / 3}}{(3 a+a+h)^{1 / 3}-(4(a+h))^{1 / 3}}$

$=\lim _{h \rightarrow 0} \frac{(3 a)^{1 / 3}\left(1+\frac{2 h}{3 a}\right)^{1 / 3}-(3 a)^{1 / 3}\left(1+\frac{h}{a}\right)^{1 / 3}}{(4 a)^{1 / 3}\left(1+\frac{h}{4 a}\right)^{1 / 3}-(4 a)^{1 / 3}\left(1+\frac{h}{a}\right)^{1 / 3}}$

$=\lim _{h \rightarrow 0}\left(\frac{3^{1 / 3}}{4^{1 / 3}}\right)\left[\frac{\left(1+\frac{2 h}{9 a}\right)-\left(1+\frac{h}{3 a}\right)}{\left(1+\frac{h}{12 a}\right)-\left(1+\frac{h}{3 a}\right)}\right]$

$=\left(\frac{3}{4}\right)^{1 / 3} \frac{\left(\frac{2}{9}-\frac{1}{3}\right)}{\left(\frac{1}{12}-\frac{1}{3}\right)}=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{8-12}{3-12}\right)$

$=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{-4}{-9}\right)=\frac{4^{1-\frac{1}{3}}}{3^{2-\frac{1}{3}}}=\frac{4^{2 / 3}}{3^{5 / 3}}$

$=\frac{(8 \times 2)^{1 / 3}}{(27 \times 9)^{1 / 3}}=\frac{2}{3}\left(\frac{2}{9}\right)^{1 / 3}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.