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How to show that $$ \lim_{x\to 1}\frac{x^3-1}{x-1}=3? $$ I tried to solve but couldn't. Please help me.

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    $\begingroup$ Did you try factoring the polynomial in the numerator? or long division to simplify the quotient? $\endgroup$
    – TravisJ
    Sep 8, 2015 at 17:19

5 Answers 5

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If $x\neq 1$, then (this is a geometric sum) $$ 1+x+x^2=\frac{x^3-1}{x-1}. $$ I guess you can proceed from here?

Edit Although I think that it is good to know and recognize geometric series, here are two other ways.

1) L'Hospital. $$ \lim_{x\to 1}\frac{x^3-1}{x-1}=\lim_{x\to 1}\frac{3x^2}{1}=3. $$

2) Since the polynomial $f(x)=x^3-1$ satisfies $f(1)=0$, we know by the factor theorem that $(x-1)$ is a factor of $f(x)$. Long polynomial division gives $$ \frac{x^3-1}{x-1}=x^2+x+1. $$ Then, the limit is easily seen to be $3$.

3) The function $f(x)=x^3$ is differentiable everywhere, with derivative $f'(x)=3x^2$. By the definition of derivative $$ 3=f'(1)=\lim_{x\to 1}\frac{x^3-1}{x-1}. $$

Choose any way, but I still like the one with geometric series best.

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  • $\begingroup$ how did you come to this step? I am not able to solve it even. $\endgroup$
    – RajSharma
    Sep 8, 2015 at 17:01
  • $\begingroup$ Honestly, I just recognized it. If one does not do that, I suggest to do a polynomial long division. $\endgroup$
    – mickep
    Sep 8, 2015 at 17:10
  • $\begingroup$ It's [arguably] easier to recognize the difference of cubes and then to factor. $\endgroup$ Sep 8, 2015 at 17:15
  • $\begingroup$ @TheChaz2.0 We are all different. I saw a geometric sum, you a difference of cubes. I like yours too. $\endgroup$
    – mickep
    Sep 8, 2015 at 17:21
  • $\begingroup$ I would have my limit approach a different x-value, if that's what you mean :) $\endgroup$ Sep 8, 2015 at 17:38
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Hint $x^3-1=(x-1)(x^2+x+1)$ and $x$ is not exactly $1$.

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  • $\begingroup$ thanks now I can solve it... $\endgroup$
    – RajSharma
    Sep 8, 2015 at 17:04
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    $\begingroup$ Gd i just gave hint so you solve it yourself and learn the way:) $\endgroup$ Sep 8, 2015 at 17:15
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    $\begingroup$ Or just practice $\endgroup$ Sep 8, 2015 at 17:16
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    $\begingroup$ I liked the comment "x is not exactly 1". Key point. $\endgroup$
    – bartgol
    Sep 8, 2015 at 19:20
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The fact that $x^3-1$ becomes $0$ when $x=1$ tells you that $x-1$ is a factor of $x^3-1$. So you have $$ x^3-1 = (x-1)(\cdots\cdots). $$ You can find the other factor by long division, then cancel $x-1$ from the numerator and the denominator. After that it's easy.

Basic fact from algebra. If you plug a number (e.g. $4$) into a polynomial and get $0$, then $x$ minus that number (e.g. $x-4$) is a factor of that polynomial.

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Let $x = y+1$, so we want to see what happens as $y \to 0$.

Note: I almost always want to have a variable that controls a limit go to zero instead of some other value (like $1$ in this case).

\begin{align} \frac{x^3-1}{x-1} &=\frac{(y+1)^3-1}{(y+1)-1}\\ &=\frac{(y^3+3y^2+3y+1)-1}{y}\\ &=\frac{y^3+3y^2+3y}{y}\\ &=y^2+3y+3 \qquad\text{whenever }y \ne 0\\ &\to 3 \qquad\text{as } y \to 0\\ \end{align}

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

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You can use polynomial long division (https://en.wikipedia.org/wiki/Polynomial_long_division) on this problem $$(x^3-1):(x-1)=x^2+x+1$$

Then taking the limit is reduced to setting $x=1$ into $x^2+x+1$.

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