Finding common denominator of a polynomial

Question

I am solving a limit and am stuck with finding the common denominator. This is the limit that I have:

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

Since $x \to 1$, the limit is $\frac{1}0 - \frac{3}0$ - or $\infty - \infty$.

So, I think the next thing I need to do is find the common denominator and continue from there on.

I would very much appreciate an answer. Thank you all and have a nice day.

$1-x^3=(1-x)(1+x+x^2)$. Therefore a common denominator can be $1-x^3$. The sum can be written as $\frac{1+x+x^2-3}{1-x^3}$. — Ginna, Jul 23 '17 at 21:45
Yes Ginna, thank you. I got your result, but what can I factor next? — Harvey Ross, Jul 23 '17 at 22:06
The numerator became $x^2+x-2=(x-1)(x+2)$. Therefore the fraction is equal to $\frac{x+2}{1+x+x^2}$. — Ginna, Jul 23 '17 at 22:19

score 1 · Answer 1 · answered Jul 23 '17 at 21:44

1

Hint:

This might help you.

$$1-x^3= (1-x)(1+x+x^2)$$

answered Jul 23 '17 at 21:44

Siong Thye Goh

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Thank you for the answer. Now I get 1+x+x^2-3/(1-x)^3. What can I factor out next? – Harvey Ross Jul 23 '17 at 22:03
your numerator become $x+x^2-2=(x-1)((x+2)$ now you can cancelled $(x-1)$ from both numerator and denominator. – Siong Thye Goh Jul 24 '17 at 00:05

score 1 · Answer 2 · answered Jul 23 '17 at 23:03

Hint: $\;1-x^3=(1-x)(1+x+x^2)\,$, then by partial fraction decomposition:

$$ \frac{1}{1-x^3} = \frac{1}{3}\left(\frac{1}{1-x}+\frac{x+2}{1+x+x^2}\right) $$

It follows that:

$$\require{cancel} \frac{1}{1-x} - \frac{3}{1-x^3} = \cancel{\frac{1}{1-x}} - \left(\cancel{\frac{1}{1-x}} + \frac{x+2}{1+x+x^2}\right) = - \,\frac{x+2}{1+x+x^2} $$

Dr. Sonnhard Graubner · Answer 3 · 2017-07-23T22:18:55.613

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HINT: note that $$1-x^3=- \left( -1+x \right) \left( {x}^{2}+x+1 \right) $$ and $$\frac{1}{1-x}-\frac{3}{1-x^3}=-{\frac {x+2}{{x}^{2}+x+1}}$$ $$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{-x^3+3x-2}{(1-x)(1-x^3}=\frac{-(x+2)(x-1)^2}{(1-x)(1-x^3)}$$

edited Jul 23 '17 at 22:18

answered Jul 23 '17 at 21:46

Dr. Sonnhard Graubner

95,283

Thank you for the answer. How did you get - (x+2)? – Harvey Ross Jul 23 '17 at 22:14

score 0 · Accepted Answer · answered Jul 23 '17 at 22:22

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$$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{1}{1-x}-\frac{3}{(1-x)(x^2+x+1)}=\frac{1}{1-x}\left(1-\frac{3}{x^2+x+1}\right)=\frac{1}{1-x}\left(\frac{x^2+x-2}{x^2+x+1}\right)=\frac{1}{1-x}\left(\frac{(x-1)(x+2)}{x^2+x+1}\right)=\frac{-(x+2)}{x^2+x+1}$$

answered Jul 23 '17 at 22:22

dromastyx

2,796

Thank you very much. Your answer was the most straightforward for me to understand. – Harvey Ross Jul 24 '17 at 11:44
@HarveyRoss I'm glad it helped :) – dromastyx Jul 24 '17 at 11:46

Finding common denominator of a polynomial

4 Answers4