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 :
Correct Option: , 2
Solution:
$\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}} \quad\left[\frac{0}{0}\right.$ case $]$
Apply L'Hospital rule
$\lim _{x \rightarrow a} \frac{\frac{1}{3}(a+2 x)^{-2 / 3} \cdot 2-\frac{1}{3} \cdot(3 x)^{-2 / 3} \cdot 3}{\frac{1}{3}(3 a+x)^{-2 / 3} \cdot-\frac{1}{3}(4 x)^{-2 / 3} \cdot 4}$
$=\frac{\frac{1}{3}(3 a)^{-2 / 3} \cdot(2-3)}{\frac{1}{3}(4 a)^{-2 / 3} \cdot(1-4)}=\frac{3^{-2 / 3}}{4^{-2 / 3}} \cdot \frac{1}{3}$
$=\frac{2^{4 / 3}}{9^{1 / 3}} \cdot \frac{1}{3}=\frac{2}{3} \cdot\left(\frac{2}{9}\right)^{1 / 3}$