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I encountered an issue for finding the limit for a function, one small step I just can't get. I know that this is classic case for L'Hospital's however our calculus course didn't reach it and I can not use it for this one.

$ 2.3 .m,n\in\mathbb{N}\quad\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}$

$x\neq1$

\begin{aligned}\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x^{m}-1^{m}}{x^{n}-1^{n}}=\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}=\\ \underset{x\rightarrow1}{\lim}\frac{\left(m-1\right)x^{m-1}}{\left(n-1\right)x^{n-1}}=\underset{x\rightarrow1}{\lim}\frac{mx^{m-1}-x^{m-1}}{nx^{n-1}-x^{n-1}} \end{aligned}

And here I stuck... know I need to reach $\frac{m}{n}$ but how can I proceed...?!!? nothing to cancel out here...

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  • $\begingroup$ Hi that's what I did there already using the sigma notation :$x^{m}-y^{m}=(x-y)\sum_{i=0}^{m-1}x^{i}\cdot y^{m-i-1}$ $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:02
  • $\begingroup$ How did you go from the end of the first line to the second? $\endgroup$ – HEXQ Mar 23 '16 at 21:09
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    $\begingroup$ The beginning of the last line is no more indeterminate, but you wrongly calculated the number of terms. From $i=0$ to $m-1$, there are $m$ of them. $\endgroup$ – Bernard Mar 23 '16 at 21:10
  • $\begingroup$ You are so close. At this point you can set x = 1. However, you have an error in the summation of the series, which is why you are not arriving at the desired result. $\endgroup$ – Doug M Mar 23 '16 at 21:26
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Another approach is along the lines of

\begin{align} L & = \lim_{x \to 1} \frac{x^m-1}{x^n-1} \\ & = \lim_{h \to 0} \frac{(1+h)^m-1}{(1+h)^n-1} \\ & = \lim_{h \to 0} \frac{1+mh+O(h^2)-1}{1+nh+O(h^2)-1} \\ & = \lim_{h \to 0} \frac{mh+O(h^2)}{nh+O(h^2)} \\ & = \frac{m}{n} \end{align}

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  • $\begingroup$ Very quick way... $\endgroup$ – imranfat Mar 23 '16 at 21:06
  • $\begingroup$ @imranfat: The division by $x-1$ is very slick, though, I must say. This is a different approach that basically does L'Hôpital without explicitly invoking L'Hôpital. $\endgroup$ – Brian Tung Mar 23 '16 at 21:09
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It seems you are aware that the numerator and denominator factor. Check the upper limit on the sigma in the denominator, and also check the conclusion that $$\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}= \underset{x\rightarrow1}{\lim}\frac{\left(m-1\right)x^{m-1}}{\left(n-1\right)x^{n-1}}.$$

The left hand side is correct (apart from the upper limit in the denominator), but the right is not. It should read $$\underset{x\rightarrow1}{\lim}\frac{\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\sum_{i=0}^{n-1}x^{j}\cdot1^{n-j-1}}=\frac{m}{n}.$$

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  • $\begingroup$ Thank you very much however could you explain why this is the solution for the sigma quotients ? $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:09
  • $\begingroup$ As I know $\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}=(m-1)x^{m-1}$ why is it m,n and not according to the boundaries of the summing $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:10
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    $\begingroup$ @user313448 No, this is false. $\sum_{i=0}^{m-1}x^i1^{m-i-1}=x^{m-1}+x^{m-2}+x^{m-3}+...+x+1$. When $x$ is close to 1, then we are just adding together $m$ things that are all close to 1, so the sum is about $m$. In the limit, it is exactly $m$. $\endgroup$ – Alex S Mar 23 '16 at 21:14
  • $\begingroup$ Thanks so much Alex! that was the missing part for my understanding! $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:15
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You can factor $$x^m-1 = (x-1)(x^{m-1} + x^{m-2} + \dots + x + 1)$$.

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  • $\begingroup$ How does factoring works on this notation ?$x^{m}-y^{m}=(x-y)\sum_{i=0}^{m-1}x^{i}\cdot y^{m-i-1}$ $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:05
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$$\frac{x^m-1}{x^n-1}=\frac{(x-1)(1+x+x^2+\cdots +x^{m-1})}{(x-1)(1+x+x^2+\cdots +x^{n-1})}$$ The numerator has $m$ terms and the denominator has $n$.

Applying the limit, we get $$\lim_{x \to 1}\frac{x^m-1}{x^n-1}=\frac{m}{n}$$

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  • $\begingroup$ How does factoring works on this notation ?$x^{m}-y^{m}=(x-y)\sum_{i=0}^{m-1}x^{i}\cdot y^{m-i-1}$ $\endgroup$ – Pavel Penshin Mar 23 '16 at 21:05
  • $\begingroup$ $$x^m-y^m=(x-y)(x^{m-1}y^0+x^{m-2}y^1+\cdots+x^0y^{m-1})$$ $\endgroup$ – Nikunj Mar 23 '16 at 21:07
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Once you factor $x-1$ from the numerator and denominator, simply cancel them out, then substitute $1$. This yields

$$\dfrac{\sum_{i=0}^{m-1}1}{\sum_{i=0}^{n-1}1}$$

Since each term in the sum is now constant, this reduces to simple multiplication. There are $m$ $1$'s in the numerator and $n$ $1$'s in the denominator. So the result is simply $\frac{m(1)}{n(1)}=\frac mn$.

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$\lim_{x\to 1} \dfrac{x^m - 1}{x^n - 1} = \lim_{x\to 1} \dfrac{(x-1)(x^{m - 1} + x^{m - 2} + x^{m - 3} + ... + 1)}{(x-1)(x^{n - 1} + x^{n - 2} + x^{n - 3} + ... + 1)} = \lim_{x\to 1} \dfrac{(x-1)\sum_{i=1}^{m} x^{m-i}}{(x-1)\sum_{j=1}^{n} x^{n-j}} = \lim_{x\to 1} \dfrac{\sum_{i=1}^{m} x^{m-i}}{\sum_{j=1}^{n} x^{n-j}} = \dfrac{\sum_{i=1}^{m} 1^{m-i}}{\sum_{j=1}^{n} 1^{n-j}} = \dfrac{\sum_{i=1}^{m} 1}{\sum_{j=1}^{n} 1} = \dfrac{m}{n}$

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\begin{aligned}\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1} &=\underset{x\rightarrow1}{\lim}\frac{x^{m}-1^{m}}{x^{n}-1^{n}}\\ &=\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}\\ &=\underset{x\rightarrow1}{\lim}\frac{\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\sum_h{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}\\ &=\dfrac mn \end{aligned}

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Define $\displaystyle L(m,n) = \underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}.$ You wish to find the value of $L(m,n)$ for $m,n\in\mathbb N$.

It’s easy to see that $L(k,k)=1$, and if $L(m,n)$ exists and is nonzero, then $L(m,n)=\dfrac{1}{L(n,m)}$. Now notice that

$$L(m+1,n)=\underset{x\rightarrow1}{\lim}\frac{x^{m+1}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x\left(x^{m}-1\right)+(x-1)}{x^{n}-1}=L(m,n)+L(1,n),$$

so $L[m,n]$ satisfies the recurrence relation

$L[m,n]={\Big\{} \begin{array}{cc} \ & \begin{array}{cc} 1 & m=n \\ \dfrac{1}{L[n,m]} & m<n \\ L(m-1,n)+L(1,n) & m>n \end{array} \end{array} $

This recurrence relation has a unique solution $L[m,n]=\dfrac{m}{n}$.

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