Mathematics - Limits Question with Solution | TestHub

At $x=a$, limit is of $\frac{0}{0}$ form.

So, applying L'Hospital rule, we get

$\underset{x\to a}{\lim }\frac{{\left(a+2x\right)}^{{\displaystyle \frac{1}{3}}}-{\left(3x\right)}^{\frac{1}{3}}}{{\left(3a+x\right)}^{\frac{1}{3}}-{\left(4x\right)}^{\frac{1}{3}}}=\underset{x\to a}{lim}\frac{{\displaystyle \frac{1}{3}{\left(a+2x\right)}^{-\frac{2}{3}}.2-\frac{1}{3}.{\left(3x\right)}^{-\frac{2}{3}}.3}}{\frac{1}{3}{\left(3a+x\right)}^{-\frac{2}{3}}.-\frac{1}{3}.{\left(4x\right)}^{-\frac{2}{3}}.4}$

$=\frac{{\displaystyle \frac{1}{3}{\left(3a\right)}^{-\frac{2}{3}}.\left(2-3\right)}}{\frac{1}{3}{\left(4a\right)}^{-\frac{2}{3}}.\left(1-4\right)}=\frac{3^{-\frac{2}{3}}}{4^{-\frac{2}{3}}}.\frac{1}{3}$

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