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Given: $\lim_{x\rightarrow 1}\frac{x-1}{x^{n}-1}, x \in \mathbb{R}$

Because you cannot really get the limit with the current given function, I have used L'Hôpitals rule.

$f(x) = x-1$

$f'(x) = 1$

$g(x) = x^n-1$

$g'(x) = nx^{n-1}$

So we got:

$\lim_{x\rightarrow 1}\frac{1}{nx^{n-1}} = 0$ for $n>1$

What confuses me much is that I don't know anything about $n$. So can I just do it like that and define $n$ myself? If it is correct, would it be better if I'd write "for any large n" instead of $n>1$ ?

Edit: The correct solution is $1/n$, not $0$!

Martin Sleziak
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berndgr
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2 Answers2

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Almost everything you did (as of the time I write this) is perfectly fine. The answer is $1/n$, not $0.$ Also, I don't understand what you mean by "define $n$ myself." You don't need to know anything about $n$ to take the derivative of $x^n-1$, as long as $n$ is a nonzero constant and not another variable (in the sense that $x$ is a variable).

Note also that if $n$ is known to be a positive integer, then you can factor $x^n-1$ to get $$ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x^2 + x + 1) $$

Then you can cancel the $x-1$ from the numerator and denominator. And because there are $n$ terms in the second factor on the RHS above, you'll get the same answer of $1/n.$

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    You actually do need to know that $n\neq 0$, otherwise you have $x^0-1=1-1=0$ in a place where it shouldn't be. – Hirshy Jul 20 '16 at 13:19
  • Thank you! To be honest, I don't like the way where you can form $x^{n}-1$ to something else and then continue. L'Hôspital works as well, right? I saw my mistake and understood why the solution is $1/n$ now. – berndgr Jul 20 '16 at 13:20
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    @Hirshy, thanks, updated. –  Jul 20 '16 at 13:21
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    @berndgr, yes, l'Hôpital is a valid method here. Just FYI, what you call "form $x^n - 1$ to something else" is just factoring, just like how, for example, $x^2 + 5x + 6$ can be factored into $(x+3)(x+2)$. Please trust me when I say it's very important that you be comfortable with factoring before you advance much further in your math studies in general. –  Jul 20 '16 at 13:23
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    @berndgr one can use L'Hospital here, yes. But: L'Hospital is quite a strong result one shouldn't use without care (see here for an example). What basically happens in tilper's manipulation is just the formula for a (finite) geometric sum which I'd consider to be a very important and standard formula to know. – Hirshy Jul 20 '16 at 13:23
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$$\frac{x-1}{x^n-1}=\frac 1{\dfrac{x^n-1}{x-1}}. $$ Now the denominator is the rate of variation of the function $x^n$ at $x=1$, hence its limit is the derivative $nx^{n-1}$ at $x=1$. So $$\frac{x-1}{x^n-1}\xrightarrow[x\to 1]{}\frac 1n. $$

Bernard
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