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

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

Martin Sleziak
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Casi
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    Do you mean $$\frac{x^4-1}{x^3-1}?$$ – Alex Becker Aug 30 '12 at 00:31
  • Yes. My Professer wasn't very clear on how to find the limit and I am very confused – Casi Aug 30 '12 at 00:33
  • 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. – Eric Naslund Aug 30 '12 at 00:38

6 Answers6

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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 $1$.

André Nicolas
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  • Thanks, but I don't think that's method I'm suppose to be using – Casi Aug 30 '12 at 00:38
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    @Casi: If you are expected at this stage to do numerical experimentation by calculator followed by a guess, then the answer by DrKW is the most appropriate one. – André Nicolas Aug 30 '12 at 00:46
  • It is frequently useful to know that $x^n-1$ is divisible by $x-1$. – MJD Aug 30 '12 at 01:03
  • 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. – Did Aug 30 '12 at 09:14
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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.

Kris Williams
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  • I have tried that however when I insert the numbers into the equation, I find answers that doesn't match the ones on the answer sheet my professor gave me. – Casi Aug 30 '12 at 00:37
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    @Casi I mean this in the nicest way possible. Make sure you are putting it in your calculator correctly. I noticed in the original posting you wrote x^4-1 / x^3-1, but your calculator needs more parentheses $(x^4-1)/(x^3-1)$. Otherwise, post what you have and I will try to help. – Kris Williams Aug 30 '12 at 00:42
  • (1.1^4−1)/(1.1^3−1)=1.402 (1.01^4-1)/(1.01^3-1)=1.34 (1.001^4-1)/(1.001^3-1)=1.334 – Casi Aug 30 '12 at 00:46
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    This looks fine, keep going!! It can take awhile until you see the pattern. This is why we have better methods for finding limits exactly, as described in other people's answers. – Kris Williams Aug 30 '12 at 00:50
  • The link you posted gave me the same answers I found which is strange because the sheet gave me numbers like 2.0572, 2.149 – Casi Aug 30 '12 at 00:53
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    I would say the sheet is wrong if it supposed to be about this problem. – Kris Williams Aug 30 '12 at 01:05
  • There must be a mistake somewhere. The limit of $(x^4-1)/(x^3-1)$ as $x$ tends to $1$ is $4/3 = 1.33333...$ – Fly by Night Aug 30 '12 at 01:14
  • I see. So then, based on the answers I found the limit would be 2 with F(1) being undefined? – Casi Aug 30 '12 at 01:37
  • @Casi: No, your numbers are 1.402, 1.34, 1.334, ... Those don't seem to be approaching 2. Try a few more and see what happens. (But you're right that $f(1)$ is undefined.) – Théophile Aug 30 '12 at 01:38
  • How about 1.3? If I add more zeroes to x, the answer gets dragged out like 1.33334 or something similiar – Casi Aug 30 '12 at 01:52
  • With $x=1.0000000000000001$ the following expression $(x^4-1)/(x^3-1)$ returns Nan in Octave, but most of the other formula here return 1.3333... – copper.hat Aug 30 '12 at 02:36
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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. – DonAntonio Aug 30 '12 at 02:56
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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.

copper.hat
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  • Your answer and André’s would be my recommendations, assuming that the problem is from early in Calc I; I really don’t like the numerical approach for anything but illustrating the notion of a limit. – Brian M. Scott Aug 30 '12 at 00:59
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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 :-).) – copper.hat Aug 30 '12 at 01:07
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Applying L'Hôpital's rule makes it quite straightforward.

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

johnny
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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.)

Michael Hardy
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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 ($\frac{0}{0}$),

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

Kirk Boyer
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