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

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The limit to evaluate is $$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$ without using l'Hôpital's rule. The discussion includes various approaches to simplify the expression as x approaches 1, focusing on algebraic manipulation and series expansion. Members successfully found the limit, with contributions highlighting the importance of recognizing indeterminate forms and applying appropriate mathematical techniques. The correct solutions were provided by several members, showcasing collaborative problem-solving. The limit can be evaluated through careful analysis of the behavior of the terms involved as x approaches 1.
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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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