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Given: $$ \lim_{x \to 1} f(x) = \frac3{1-x^3} - \frac1{1-x} $$

I first used the difference of cubes to get $$ \lim_{x \to 1} f(x) = \frac3{(1-x)(1+x+x^2)} - \frac1{1-x} $$

Then multiplied each term by $(1-x)$, cancelling $(1-x)$ in the first term, and making the second term $1$ $$ \lim_{x \to 1} f(x) = \frac3{(1+x+x^2)} - 1 $$

After substitution $$ \lim_{x \to 1} f(x) = \frac33 - 1 $$

Which equates to $0$. But I know the correct answer is $1$. Where did I slip up?

antgel
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    What makes you think you can cancel $(1-x)$ ? To add/subtract fractions, you need to get common denominator first : $$\begin{align} \\lim_{x \to 1} f(x) &= \lim_{x \to 1} \frac3{(1-x)(1+x+x^2)} - \frac{1}{1-x}\~\&= \lim_{x \to 1} \frac3{(1-x)(1+x+x^2)} - \frac{(1+x+x^2)}{(1-x)(1+x+x^2)}\end{align}$$ – AgentS Sep 02 '14 at 13:28
  • Thanks, I get it now. I made a mistake in my intended calculation, but I was overcomplicating matters anyway. – antgel Sep 02 '14 at 13:46

2 Answers2

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A start: Bring the full expression to a common denominator by multiplying top and bottom of the second term by $1+x+x^2$. After minor simplification, we get $$\frac{2-x-x^2}{(1-x)(1+x+x^2)}$$ Now note that $2-x-x^2=(2+x)(1-x)$.

André Nicolas
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  • When you say "Now note that $2-x-x^2=(2+x)(1-x)$.", was that your intuition? Because it's not intuitive to me at present. If not, what was your reasoning? – antgel Sep 02 '14 at 14:38
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    The polynomial $2-x-x^2$ has root $x=1$. So $x-1$ (or equivalently $1-x$) must divide $2-x-x^2$. If we had started with $\frac{4}{1-x^3}$ instead of $\frac{3}{1-x^3}$, we would not have this "lucky" break. But a similar common denominator calculation would show us that the limit does not exist. – André Nicolas Sep 02 '14 at 14:50
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You can't multiply each term by $(1-x)$- rather you should factor it out to get:

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

Now, cancel, and take the limit- the answer shall be $1$

voldemort
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