Limit of particular function.

Question

I need to find the limit of the following function: $$\lim\limits_{x\to 1}\frac{1}{3(x^{1/3}-1)}-\frac{1}{2(x^{1/2}-1)}.$$

Answer 1

Let $x=u^6$. Notice that as $x\to1$, $u\to1$, so we have

$$\begin{align}\lim_{x\to1}\frac1{3(x^{1/3}-1)}-\frac1{2(x^{1/2}-1)}&=\lim_{u\to1}\frac1{3(u^2-1)}-\frac1{2(u^3-1)}\\&\vphantom{\cfrac11}=\lim_{u\to1}\frac1{u-1}\left(\frac1{3(u+1)}-\frac1{2(u^2+u+1)}\right)\\&\vphantom{\cfrac11}=\lim_{u\to1}\frac1{u-1}\left(\frac{2(u^2+u+1)}{6(u+1)(u^2+u+1)}-\frac{3(u+1)}{6(u+1)(u^2+u+1)}\right)\\&\vphantom{\cfrac11}=\lim_{u\to1}\frac1{u-1}\left(\frac{2u^2-u-1}{6(u+1)(u^2+u+1)}\right)\\&\vphantom{\cfrac11}=\lim_{u\to1}\frac1{u-1}\left(\frac{(u-1)(2u+1)}{6(u+1)(u^2+u+1)}\right)\\&\vphantom{\cfrac11}=\lim_{u\to1}\frac{2u+1}{6(u+1)(u^2+u+1)}\\&\vphantom{\cfrac11}=\frac3{6\times2\times3}\\&\vphantom{\cfrac11}=\frac1{12}\end{align}$$

Answer 2

Since we can assume $x>0$, we can do the substitution $x=t^6$, with $t>0$, so the limit becomes $$ \lim_{t\to1}\left(\frac{1}{3(t^2-1)}-\frac{1}{2(t^3-1)}\right) $$ Can you take from here?