<s>[INST] Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I do next? Remember, I can't use L'Hôpital.
 [/INST] $$\lim_{x\to1}\frac{\ln x}{x-1}=1$$ is a basic limit from high school, and expresses the logarithm is differentiable at $x=1$ and has derivative equal to $1$.
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