Without using L'Hopital's rule, evaluate $\lim_{x\to6} \frac{x^3−3x^2-x-102}{x-6}$. Can anyone please help?
Thanks
2026-03-29 10:59:57.1774781997
Solving limit without L'Hôpital (without square root)
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By factorizing $x^3-3x^2-x-102$, we get $$x^3-3x^2-x-102=(x-6)(x^2+3x+17)$$ Therefore, the limit becomes: $$\lim_{x\rightarrow 6}\dfrac{x^3-3x^2-x-102}{x-6}=\lim_{x\rightarrow 6}\dfrac{(x-6)(x^2+3x+17)}{x-6}\\=\lim_{x\rightarrow 6} (x^2+3x+17)=6^2+3(6)+17=\boxed{71}$$