How can we evaluate this limit without using l'Hôpital's rule?

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The limit evaluation discussed is $$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$. The solutions provided by members MarkFL, anemone, Sudharaka, and soroban successfully demonstrate methods to evaluate this limit without employing l'Hôpital's rule. Sudharaka's solution involves algebraic manipulation and the application of Taylor series expansions around the point of interest, yielding a definitive limit value.

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Evaluate the following limit without using l'Hôpital's rule.

$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$.

Hint:
Use the substitution $x=u^6$.
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) Sudharaka
4) soroban

Solution (from Sudharaka):
Substitute \(x=u^6\) and we get,

\begin{eqnarray}

\lim_{x\to 1}\left(\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}\right)&=&\lim_{u\to 1}\left(\frac{1}{2(1 - u^3)} - \frac{1}{3(1 - u^2)}\right)\\

&=&\frac{1}{6}\lim_{u\to 1}\left[\frac{1-3u^2+2u^3}{(1 - u^2)(1-u^3)}\right]\\

&=&\frac{1}{6}\lim_{u\to 1}\left[\frac{(1-u)^2 (1+2u)}{(1 - u^2)(1-u^3)}\right]\\

&=&\frac{1}{6}\lim_{u\to 1}\left[\frac{1+2u}{(1+u)(1+u+u^2)}\right]\\

&=&\frac{1}{12}

\end{eqnarray}
 

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