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In trying to evaluate the following limit: $$\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$$

I am getting the indefinite form of: $$\frac{1}{\mbox{undefined}}-\frac{3}{\mbox{undefined}}$$ What would be the best solution to evaluating this limit?

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  • $\begingroup$ You rather get the indefinite form $\frac{1}{0} - \frac{3}{0}$. $\endgroup$ – md2perpe Aug 16 '17 at 20:36
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Hint. Note that for $x\not=1$, $$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{x^2+x+1-3}{(1-x)(x^2+x+1)} =\frac{(x+2)(x-1)}{(1-x)(x^2+x+1)}=-\frac{x+2}{x^2+x+1}. $$

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  • $\begingroup$ @Omari Celestine Any further doubt? $\endgroup$ – Robert Z Aug 16 '17 at 18:39
  • $\begingroup$ I now understand. Thank you. $\endgroup$ – Omari Celestine Aug 18 '17 at 15:39
  • $\begingroup$ @Omari Celestine Well done!! $\endgroup$ – Robert Z Aug 18 '17 at 15:42
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Hint: $$\frac{1}{1-x}-\frac{3}{(1-x)(1+x+x^2)}=\frac{x^2+x-2}{1-x^3}.$$ Use L'Hospital's rule to compute the limit.

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    $\begingroup$ and $$x\neq 1$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 16 '17 at 18:33
  • $\begingroup$ Thanks @Dr.SonnhardGraubner. $\endgroup$ – Math Lover Aug 16 '17 at 18:35
  • $\begingroup$ I was not sure how to transform the equation. Could you explain how you got to the final equation? $\endgroup$ – Omari Celestine Aug 16 '17 at 18:37
  • $\begingroup$ @OmariCelestine Note that $1-x^3=(1-x)(1+x+x^2)$. Rest is just the addition of two fractions. $\endgroup$ – Math Lover Aug 16 '17 at 18:39
  • $\begingroup$ ok. thanks much. $\endgroup$ – Omari Celestine Aug 16 '17 at 18:50

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