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Evaluate $\lim_{x \to 1} (\frac{1}{1-x} - \frac{3}{1-x^3}) $

We have $$(1-x^3) = (1-x)(1+x+x^2) \implies \frac{1}{1-x} = \frac{(1+x+x^2)}{(1-x^3)}$$

So we have $$\frac{1}{1-x} - \frac{3}{1-x^3} = \frac{(1+x+x^2)}{(1-x^3)} - \frac{3}{1-x^3} = \frac{x^2 + x - 2}{1-x^3} =$$ $$ \frac{(x+2)(x-1)}{(1-x)(1+x+x^2)} = - \frac{(x+2)}{(1+x+x^2)}$$

Now, we can safely evaluate

$$\lim_{x \to 1} - \frac{x+2}{1+x+x^2} = - \frac{1+2}{1+1+1^2} = - \frac{3}{3} = - 1$$

Is this correct ? They don't give out solutions to this problem

wengen
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    There's one small mistake: when you cancel the factors $x-1$ and $1-x$, a negative sign should appear since one is the negative of the other. Once you insert that negative sign in the rest of the calculation, this looks great (and the best way to do this problem)! – Greg Martin Nov 05 '23 at 19:10
  • https://math.stackexchange.com/questions/2297504/find-lim-x-to-1-fracp1-xp-fracq1-xq – lab bhattacharjee Nov 05 '23 at 19:46

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We can write your question as $\lim_{x \to 1}\frac{-3+1+x+x^2}{1-x^3}$. Let $L=\lim_{x \to 1}\frac{-2+x+x^2}{1-x^3}$. By L'Hospital, we we see $L=\lim_{x \to 1}\frac{2x+1}{-3x^2}$. Hence $L=-\frac{3}{3}$.

Surajit131
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