Evaluating the limit $\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$

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?

• You rather get the indefinite form $\frac{1}{0} - \frac{3}{0}$. – md2perpe Aug 16 '17 at 20:36

Hint. Note that for $x\not=1$, $$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{x^2+x+1-3}{(1-x)(x^2+x+1)} =\frac{(x+2)(x-1)}{(1-x)(x^2+x+1)}=-\frac{x+2}{x^2+x+1}.$$

• @Omari Celestine Any further doubt? – Robert Z Aug 16 '17 at 18:39
• I now understand. Thank you. – Omari Celestine Aug 18 '17 at 15:39
• @Omari Celestine Well done!! – Robert Z Aug 18 '17 at 15:42

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.

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