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.
 A: 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
$$253\dfrac{12}{-276+782}=6$$
A: 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 $6$.
A: 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.
A: 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, $$\lim_{x\to 1}\left[\frac{n}{1-x^n}+\frac{1}{1-x}\right]=\frac{n-1}{2}.$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}\require{cancel}
&\color{#f00}{\lim_{x \to 1}\pars{{23 \over 1 - x^{23}} - {11 \over 1 - x^{11}}}} =
\lim_{x \to 1}\bracks{{1 \over 1-x}
\pars{{23 \over \sum_{k = 0}^{22}x^{k}} - {11 \over \sum_{k = 0}^{10}x^{k}}}}
\\[5mm] = &\
\lim_{x \to 1}
\bracks{23\,{\sum_{k = 1}^{22}k\,x^{k - 1} \over
\pars{\sum_{k = 0}^{22}x^{k}}^{2}} -
11\,{\sum_{k = 1}^{10}k\,x^{k - 1} \over \pars{\sum_{k = 0}^{10}x^{k}}^{2}}}
\qquad\pars{~By\ L'H\hat{o}pital\ Rule}
\\[5mm] = &\
23\,{\sum_{k = 1}^{22}k \over \pars{\sum_{k = 0}^{22}1}^{2}} -
11\,{\sum_{k = 1}^{10}k \over \pars{\sum_{k = 0}^{10}1}^{2}}
\\[5mm] = &\
\underbrace{\cancel{23}\,{22\cancel{\pars{22 + 1}}/2 \over \cancel{23^{2}}}}_{\ds{=\ 11}}\ -\
\underbrace{\cancel{11}\,{10\cancel{\pars{10 + 1}}/2 \over \cancel{11^{2}}}}_{\ds{=\ 5}}\ = \ 11 - 5 = \color{#f00}{6}
\end{align}
A: This is definitely the best method to solve this question:
$$Let\space L = \lim_{x \to 1} \left(\frac{p}{1-x^{p}}-\frac{q}{1-x^{q}} \right)$$$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ Replace $x$ by $\frac{1}{x}$
$$\space L = \lim_{{x} \to 1} \left(\frac{p}{1-{\frac{1}{x}}^{p}}-\frac{q}{1-{\frac{1}{x}}^{q}} \right)$$
$$\space L = \lim_{{x} \to 1} \left(\frac{-px^p}{1-{{x}}^{p}}-\frac{-qx^q}{1-{{x}}^{q}} \right)$$
Adding both the equations we get $2L={p-q}$
And we are done
A: $$ =\lim_{x \to 1} \frac{23 - 23x^{11} -11 + 11x^{23}}{1 - x^{11} - x^{23} + x^{34}} = \lim_{x \to 1} \frac{-23\cdot 11 x^{10} + 11\cdot 23 x^{22}}{-11x^{10} - 23x^{22} + 34x^{33}} =$$$$= \lim_{x \to 1} \frac{-23\cdot 11 \cdot 10 x^9 + 11\cdot 23 \cdot 22x^{21}}{-11\cdot 10 \cdot x^9 - 23\cdot 22x^{21} + 34\cdot 33 x^{32}} = \frac {3036}{506} = 6 $$
where we used Hopital twice
A: $\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$
$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$
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  $= 6$
A: 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}
A: 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}
A: Dumping my answer here to a different but similar question which was labeled as a duplicate and would no longer accept the answer that I had already typed.
$\lim\limits_{t\to1}\frac{5}{1-t^5}-\frac{2}{1-t^2}=$
$\lim\limits_{t\to1}\frac{5}{(1-t)(1+t+t^2+t^3+t^4)}-\frac{2}{(1-t)(1+t)}=$
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}$
Clearly the numerator evaluates to $0$ when $t=1$ so $t-1$ must be a factor.
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^4+2t^3+2t^2-3t-3}{(t-1)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^3+4t^2+6t+3}{(1+t+t^2+t^3+t^4)(1+t)}=$
$\frac{2+4+6+3}{(1+1+1+1+1)(1+1)}=\frac{15}{(5)(2)}=\frac32$
(Long division was used to obtain $\frac{2t^4+2t^3+2t^2-3t-3}{t-1}=2t^3+4t^2+6t+3$ )
