Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$

How does one evaluate the following limit? $$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$ The answer is $6$.

How does one justify this answer?

Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.

• Put the fractions over a common denominator. You get $\frac {P(x)}{Q(x)}$ where $P,Q$ are some polynomials. If the limit exists $(x-1)$ will divide both $P$ and $Q.$ Do the division. Evaluate at $1.$ if you still get an indeterminate, then do the division by $(x-1)$ as many times as necessary. – Doug M Sep 28 '16 at 20:19

Both fractions are unbounded as $x\rightarrow 1$. But if we rewrite $$\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}}=\dfrac{23(1-x^{11})-11(1-x^{23})}{(1-x^{23})(1-x^{11})}=\dfrac{11x^{23}-23x^{11}+12}{1-x^{23}-x^{11}+x^{34}}$$ we can use L'Hopital since both sides tend to 0 as $x$ tends to 1. Differentiating both sides give $$\dfrac{253x^{22}-253x^{10}}{-23x^{22}-11x^{10}+34x^{33}}=253\dfrac{x^{12}-1}{-23x^{12}-11+34x^{23}}$$ Both sides still tend to 0, so we differentiate again and get $$253\dfrac{12x^{11}}{-276x^{11}+782x^{22}}$$ which tends to $$253\dfrac{12}{-276+782}=6$$

• I didn't expect this many responses to my question, but your approach is the clearest and most standard of all of the other approaches. Also, I didn't think combining the fraction was necessary but it is. Because I see that it is much better to deal with indeterminate forms of $\frac 00$ instead of $\infty - \infty$, because $\frac 00$ prompts us to use immediately l'Hopital's rule. – user373314 Sep 28 '16 at 20:56
• @user373314 This is a very interesting comment. In France, at least what I have seen, L'Hopital's rule is hardly ever used, if even taught at all. The standard way to compute limits is by développements limités (that is, Taylor series with $O$ or $o$ notation). However, by reading american calculus books, I have discovered L'Hopital's rule is much more used there. They are mostly equivalent anyway, but this difference in teaching is reflected in questions and answers on MSE. – Jean-Claude Arbaut Sep 28 '16 at 21:04
• See also this question of mine about L'Hopital (which, coming from France, is something I may have used maybe 5 times in my life). – Clement C. Sep 28 '16 at 21:06
• @ClementC. Yes, I have seen this. Good question. L'Hopital's rule seems to have a bad reputation, however, it's interesting to see in Franklin's Advanced calculus, that even Taylor series may be derived from L'Hopital's rule. It's not bad at all, but it must be used carefully. – Jean-Claude Arbaut Sep 28 '16 at 21:13

This answer does not use L'Hopital (personal taste), only a standard identity restated below, the binomial theorem, and a straightforward Taylor expansion to first order at $0$.

Using the identity $1-x^{2n+1} = (1-x)\sum_{k=0}^{2n} x^k$, we can rewrite \begin{align*} \frac{23}{1-x^{23}} - \frac{11}{1-x^{11}} &= \frac{1}{1-x}\left(\frac{23}{\sum_{k=0}^{22}x^k} - \frac{11}{\sum_{k=0}^{10}x^k} \right)\\ &= \frac{1}{1-x}\left(\frac{23\sum_{k=0}^{10}x^k}{\sum_{k=0}^{22}x^k\sum_{k=0}^{10}x^k} - \frac{11\sum_{k=0}^{22}x^k}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k} \right)\\ &= \frac{1}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{1}{1-x}\left(23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k \right)\\ \end{align*} Let us focus on the parenthesis (the first factor converges to $\frac{1}{11\cdot 23}$ by continuity, the second is the problematic one that will be "offset" by the parenthesis).

Writing $x=1+h)$ (where we will have $h\to 0$), we get, for any fixed integer $n$, \begin{align*} \sum_{k=0}^{n}x^k &= \sum_{k=1}^{n}(1+h)^k = \sum_{k=1}^{n} \sum_{\ell=0}^k \binom{k}{\ell} h^\ell \\ &= \sum_{k=0}^{n}(1+kh +o(h)) \\ &= (n+1)+\frac{n(n+1)}{2}h +o(h) \end{align*} when $h\to 0$, as $n$ is a constant. In particular, this implies \begin{align*} 23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k &= 23\cdot 11+23\cdot \frac{11\cdot10}{2}h - 11\cdot 23-11\cdot \frac{22\cdot 23}{2}h + o(h)\\ &= 23\cdot 11\cdot (-6h) + o(h)\\ &= 23\cdot 11\cdot 6(1-x) + o(1-x) \end{align*} Overall, we thus have \begin{align*} \frac{23}{1-x^{23}} - \frac{11}{1-x^{11}} &= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{6(1-x)+o(1-x)}{1-x} \\ &= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot (6+o(1)) \xrightarrow[x\to1]{} \frac{23\cdot 11}{23\cdot 11} \cdot 6 = 6 \end{align*} as claimed.

• Darn, took much longer to type than expected. Oh, well. – Clement C. Sep 28 '16 at 20:55
• I tried an approach like yours originally but I got stuck at dealing with that $\frac 1{1-x}$. Your solution to this dilemma was simply to write $x=1+h$ (for $h > 0$)? I guess since we're taking the limit to $1$ from both sides, that must be why. – user373314 Sep 28 '16 at 21:01
• @user373314 Yes, basically. The rationale is that while it does not change anything mathematically, I am much more comfortable with standard machinery (Taylor expansions, limits, asymptotics) at $0$ that at a finite non-zero point. – Clement C. Sep 28 '16 at 21:03
• Well done. +1 . – Mark Viola Sep 28 '16 at 23:22
• Always easier to have something go to zero. – marty cohen Sep 29 '16 at 4:38

As $t\to0$, we have

$$\frac{23}{1-(1+t)^{23}}=-\frac{23}{23t+253t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+11t+O(t^2)}=-\frac{1}{t}\left(1-11t+O(t^2)\right)$$

Likewise

$$\frac{11}{1-(1+t)^{11}}=-\frac{11}{11t+55t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+5t+O(t^2)}=-\frac{1}{t}\left(1-5t+O(t^2)\right)$$

So the difference is

$$-\frac{1}{t}\left(1-11t-1+5t+O(t^2)\right)=6+O(t)$$

And your limit is $6$.

That is the same as

$$\lim_{x\to 0}\left[\frac{23}{1-(1-x)^{23}}-\frac{11}{1-(1-x)^{11}}\right]=\lim_{x\to 0}\left[\frac{23}{23x-253x^2}-\frac{11}{11-55x^2}\right]$$ (we exploited the binomial theorem and neglected terms with high order, since we can, see the comments below) or as $$\lim_{x\to 0}\frac{1}{x}\left[\frac{1}{1-11x}-\frac{1}{1-5x}\right]=\lim_{x\to 0}\left[\frac{-5+11}{(1-11x)(1-5x)}\right]=11-5=\color{red}{6}.$$ With the same approach, for any $n,m\in\mathbb{N}^+$, $$\lim_{x\to 1}\left[\frac{m}{1-x^m}-\frac{n}{1-x^n}\right]= \color{red}{\frac{m-n}{2}}.$$ In other terms, the function $f_n(x)=\frac{n}{1-x^n}$ has a simple pole at $x=1$. If we remove the contribute given by such simple pole, we are left with a holomorphic function in a neighbourhood of $x=1$. In particular, $$\lim_{x\to 1}\left[\frac{n}{1-x^n}+\frac{1}{1-x}\right]=\frac{n-1}{2}.$$

• I don't like very much to "neglect terms". There is the notation $O(x^3)$ for this purpose (or $o(x^2)$ if you prefer). More correct, and less dangerous: what happens when all previous terms cancel, you are left with $0$ while there is indeed some nonzero function of $x$? For instance, how do you know in advance that you can neglect terms of degree $>2$, but not those of degree $2$? Of course, I don't know either with the $O$ notation, but at least if I fail I will know. – Jean-Claude Arbaut Sep 28 '16 at 20:41
• @Jean-ClaudeArbaut: $$\frac{23}{23x-253x^2+o(x^2)}=\frac{23}{23x-253x^2}+o(1)$$ hence we are allowed to do that. – Jack D'Aurizio Sep 28 '16 at 20:42
• Yes, but we would also be allowed to do that at order $2$. But it would fail, and you would not know, unless you have mentally used the correct notation, but then why not write it? Bad habit, IMO – Jean-Claude Arbaut Sep 28 '16 at 20:46
• Otherwise, you may simply keep the terms with high order, then check in the last steps that they do not matter. – Jack D'Aurizio Sep 28 '16 at 20:46
• I know it works here, because even if there are terms left, they tend to $0$ and only contribute by addition, so no risk. But if you have to justify this, it will take at least as much work as if you had done things cleanly to begin with. Not worth it. And I have seen limits for which you need to keep track of terms of degree, say, $6$, so it's not a good advice, to neglect terms, when you get the same result by using a more correct notation, that does not need extra justification. – Jean-Claude Arbaut Sep 28 '16 at 20:55


• How did you justify your second equality? Because $\frac 1{1-x} = \sum_{k=0}^\infty x^k$, an infinite series. How did you get finite series in the numerator? By squaring the denominator? – user373314 Sep 29 '16 at 8:07
• @user373314 Note that $1 - x^{n} = \left(1 - x\right)\left(1 + x + x^{2} + \cdots + x^{n - 1}\right) = \left(1 - x\right)\sum_{k = 0}^{n - 1}x^{k}$ for $n = 1, 2,3,\ldots$. They are $\color{#f00}{finite\ sums}$. – Felix Marin Sep 29 '16 at 18:07

$$=\lim_{x \to 1} \frac{23 - 23x^{11} -11 + 11x^{23}}{1 - x^{11} - x^{23} + x^{34}} = \lim_{x \to 1} \frac{-23\cdot 11 x^{10} + 11\cdot 23 x^{22}}{-11x^{10} - 23x^{22} + 34x^{33}} =$$$$= \lim_{x \to 1} \frac{-23\cdot 11 \cdot 10 x^9 + 11\cdot 23 \cdot 22x^{21}}{-11\cdot 10 \cdot x^9 - 23\cdot 22x^{21} + 34\cdot 33 x^{32}} = \frac {3036}{506} = 6$$ where we used Hopital twice

$\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$

$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$

Now we could apply L'Hopitals at this point, or we can use algebra.

Using algebra, numerator and denominator both divide by $(x-1)^2$

$1-x^{11} = (1-x)\sum_\limits{i=0}^{10} x^i\\ 1-x^{23} = (1-x)\sum_\limits{i=0}^{22} x^i$

$11x^{23} -23x^{11}+ 12 = (x-1)(11 x^{22}\cdots 11x^{11} - 12x^{10}\cdots +12)\\ =(x-1)^2 (11x^{21} + 2\cdot11 x^{20} + 3\cdot11 x^{19}\cdots +12\cdot 11 x^{10} + 12\cdot 10 x^9\cdots +12\cdot 2x + 12$

Evaluated at 1.

The denominator:

$\sum_\limits{i=0}^{10} x^i = 11, \sum_\limits{i=0}^{22} x^i = 23$

the numerator:

$11 \sum_\limits{i=1}^{11} i + 12 \sum_\limits{i=1}^{11} i = (23)(11)(12)/2$

and the ratio $= 6$

Write $P_n(x)=1+x+x^2+\dots+x^n$; then our function is $$\frac{23P_{10}(x)-11P_{22}(x)}{(1-x)P_{22}(x)P_{10}(x)}$$ We can notice that $$\lim_{x\to1}P_{22}(x)P_{10}(x)=23\cdot11$$ so we just need to compute $$\lim_{x\to1}\frac{11P_{22}(x)-23P_{10}(x)}{x-1}= 11P_{22}'(1)-23P_{10}'(1)$$ by definition of derivative. Since $$P_n'(x)=1+2x+3x^2+\dots+nx^{n-1}$$ we have $$P_n'(1)=\frac{n(n+1)}{2}$$ Thus, reinserting $23\cdot11$ at the denominator, our limit is $$\frac{11\cdot22\cdot23-23\cdot10\cdot11}{2\cdot23\cdot11}= \frac{22-10}{2}=6$$

More generally, \begin{align} \lim_{x\to1}\left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right) &= \lim_{x\to1}\frac{mP_{n-1}(x)-nP_{m-1}(x)}{(1-x)P_{m-1}(x)P_{n-1}(x)} \\[6px] &= \frac{nP_{m-1}'(1)-mP_{n-1}'(1)}{mn} \\[6px] &= \frac{n(m-1)m-m(n-1)n}{2mn} \\[6px] &= \frac{m-n}{2} \end{align}

The following standard formula is well known $$\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = n = \lim_{t \to 0}\frac{(1 + t)^{n} - 1}{t}\tag{1}$$ and it appears that we can go very easily to the next step if $n$ is a positive integer and derive the formula $$\lim_{x \to 1}\frac{x^{n} - 1 - n(x - 1)}{(x - 1)^{2}} = \lim_{t \to 0}\frac{(1 + t)^{n} - 1 - nt}{t^{2}} = \frac{n(n - 1)}{2}\tag{2}$$ The simplest approach to prove $(2)$ is to use Binomial theorem. Hence we have $$x^{n} - 1 = n(x - 1) + \frac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})$$ and therefore \begin{align} \frac{n}{1 - x^{n}} &= \dfrac{n}{n(1 - x) - \dfrac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})}\notag\\ &= \frac{1}{1 - x}\left(1 - \frac{n - 1}{2}(1 - x) + o((1 - x))\right)^{-1}\notag\\ &= \frac{1}{1 - x}\left(1 + \frac{n - 1}{2}(1 - x) + o(1 - x)\right)\notag\\ &= \frac{1}{1 - x} + \frac{n - 1}{2} + o(1)\tag{3} \end{align} It follows that $$\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}} = \frac{n - m}{2} + o(1)$$ and hence $$\lim_{x \to 1}\left(\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\right) = \frac{n - m}{2}$$ and putting $n = 23, m = 11$ we get the desired limit as $6$.

The gymnastics of series division to reach $(3)$ can be avoided in another manner by using $(2)$ directly. We have \begin{align} L &= \lim_{x \to 1}\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\notag\\ &= \lim_{x \to 1}\frac{n(1 - x^{m}) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\ &= \lim_{x \to 1}\frac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\ &= \lim_{x \to 1}\dfrac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{\dfrac{(1 - x^{n})(1 - x^{m})}{(1 - x)^{2}}\cdot(1 - x)^{2}}\notag\\ &= \frac{1}{mn}\lim_{x \to 1}n\frac{1 - x^{m} - m(1 - x)}{(1 - x)^{2}} - m\frac{1 - x^{n} - n(1 - x)}{(1 - x)^{2}}\notag\\ &= \frac{1}{mn}\left(\frac{nm(1 - m)}{2} - \frac{mn(1 - n)}{2}\right)\text{ (using (2))}\notag\\ &= \frac{n - m}{2}\notag \end{align}