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What is the limit as $x \to 1$ of the function

$ f(x) = \frac{x^4-1}{x^3-1} . $

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    Related: http://math.stackexchange.com/questions/33970/finding-the-limit-of-fracqnpn-where-q-p-are-polynomials The correct change of variables, $x\rightarrow \frac{1}{x-1}$ tells us about your limit above.2012-08-30

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Note that $x^4-1=(x-1)(x^3+x^2+x+1)$ and $x^3-1=(x-1)(x^2+x+1)$.

Thus $f(x)=\frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)}.$ When $x\ne 1$, the $x-1$ terms cancel. Now we can safely let $x$ approach $1$.

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    *I don't think that's method I'm suppose to be using*... Please remind us where in your post the method you are supposed to use is specified.2012-08-30
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Generally, the "numerical method" for finding limits involves substituting numbers closer and closer to the number and seeing if we get a pattern.

In this example, we have $f(x)=\dfrac{x^4-1}{x^3-1}$. We should determine $f(1.1), f(1.01), f(0.99), f(1.001), f(0.999), \dots$ and see if a pattern develops. If these numbers approach some number, that would be our numerical estimate of the limit.

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    If you like this answer better than André's then either you don't study calculus or else you do but your teacher is trying to make some point about numerical calculations. Either way, this is one of the main reasons we do require askers to write down their background, ideas, own efforts, etc: to know how to answer them. André's answer, or Copper's, or Kirk's or Johnny's would fit in any basic calculus class.2012-08-30
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If we just divide, we get $f(x) = x+ \frac{1}{1+x+x^2}$, from which the limit easily follows, numerically or otherwise.

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    @BrianM.Scott: I agree, I'm not sure what one learns from a 'numerical approach'. (Although in secondary school, repeatedly pressing $\cos$, $\sin$, etc. on my Sinclair Cambridge calculator made me curious about fixed points :-).)2012-08-30
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Applying L'Hôpital's rule makes it quite straightforward.

$ \lim_{x \rightarrow 1}\; \frac{x^4-1}{x^3-1} = \lim_{x \rightarrow 1} \;\frac{4x^3}{3x^2} = \lim_{x \rightarrow 1} \;\frac{4}{3}x = \frac{4}{3}$

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A crucial fact from algebra should be remembered: If you plug $x=1$ into a polynomial and get $0$, that tells you that $(x-1)$ is one of the factors. (Likewise if you plug in $x=9$ and get $0$, then $(x-9)$ is one of the factors, etc.) Thus $ \frac{x^4-1}{x^3-1} = \frac{(x-1)(\cdots\cdots\cdots)}{(x-1)(\cdots\cdots\cdots)}. $ So cancel $(x-1)$ from the top and the bottom and go on from there. (To find the expressions to put in place of "$(\cdots\cdots\cdots)$", you can use long division if all else fails.)

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The most straightforward approach probably involves the fact that $x^{n}-1 = (x-1)(x^{n-1}+\dots+x+1)$, which you can verify by multiplying a few of the terms and noticing the pattern. If you rewrite the numerator and denominator using this relation, there will be cancellation of the term $(x-1)$, which is okay since $x-1=0$ only when $x$ is $1$, itself, an we are taking a limit as $x\rightarrow 1$.

Alternatively, If you have learned L'Hopital's Rule, then since "plugging in 1 for x" yields an indeterminate form ($\frac{0}{0}$),

$\lim_{x\rightarrow 1}\frac{x^{4}-1}{x^{3}-1} = \lim_{x\rightarrow 1}\frac{\frac{d}{dx}x^{4}-1}{\frac{d}{dx}x^{3}-1}$ $= \lim_{x\rightarrow 1}\frac{4x^{3}}{3x^{2}} = \lim_{x\rightarrow 1}\frac{4x}{3} = \frac{4}{3}$