The forum discussion focuses on solving the limit $\lim_{x\to1}\frac{\sin(x^{3}-1)}{x-1}$ without using L'Hôpital's rule. Participants suggest multiplying the numerator and denominator by $x^{2} + x + 1$ to simplify the expression. The key insight is recognizing that $\lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x^{3}-1} = 1$ and $\lim_{x \rightarrow 1} (x^{2} + x + 1) = 3$, leading to the final result of 3. This method effectively circumvents the need for L'Hôpital's rule while providing a clear path to the solution.
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Yankel said:Hello,
I need some help with this limit, I have no clue how to do this with the next constraint: No use of L'hopital rule...
Thank you !
\lim_{x\to1}\frac{sin(x^{3}-1)}{x-1}
chisigma said:Try multiplying numerator and denominator by $x^{2} + x + 1$...
Kind regards
$\chi$ $\sigma$
Yankel said:I thought about it, it gives me what's in the sine, however x->1 and not x->0, so I am still stuck...