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Evaluate the following limit without using L'Hospital method $$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$ My turn is $$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$ $$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$ then divide the numerator and the denominator by $$x-1$$ But I got again $$\frac{0}{0}$$

John Doe
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Hussien Mohamed
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4 Answers4

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Observe that, by the standard binomial expansion, $$ (1+h)^9-1=9h+36h^2+O(h^3),\qquad (1+h)^5-1=5h+10h^2+O(h^3) $$ then, by setting $x=1+h$, as $x \to 1$ we have $h \to 0$, we get $$ \begin{align} \lim_{x\to1}\left(\frac{9}{x^9-1}-\frac{5}{x^5-1}\right)&=\lim_{h\to0}\left(\frac{9}{(1+h)^9-1}-\frac{5}{(1+h)^5-1}\right) \\\\&=\lim_{h\to0}\left(\frac{9}{9h+36h^2+O(h^3)}-\frac{5}{5h+10h^2+O(h^3)}\right) \\\\&=\lim_{h\to0}\left(\frac{1}{h+4h^2+O(h^3)}-\frac{1}{h+2h^2+O(h^3)}\right) \\\\&=\lim_{h\to0}\frac{-2h^2+O(h^3)}{(h+4h^2+O(h^3))(h+2h^2+O(h^3))} \\\\&=-2. \end{align} $$

Olivier Oloa
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    Alternative for the penultimate step: It's $\lim_{h\to0}\frac1h(\frac1{1+4h+O(h^2)}-\frac1{1+2h+O(h^2)})$, and the two interior terms are $(1-4h+O(h^2))-(1-2h+O(h^2))=-2h+O(h^2)$ by geometric series/binomial expansion. – Steven Stadnicki Apr 21 '18 at 01:46
  • (+1) I like this more than my answer! – John Doe Apr 21 '18 at 01:48
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Dividing numerator and denominator by $x-1$ should give $$\frac{9(x^4+x^3+x^2+x+1)-5(x^8+\cdots+x+1)}{(x^{13}+\cdots+1)-(x^8+\cdots+1)-(x^4+\cdots+1)}$$As you say, this gives $0/0$. We can write it as $$\require{cancel}\frac{-5x^8-5x^7-5x^6-5x^5+4x^4+4x^3+4x^2+4x+4}{x^{13}+x^{12}+x^{11}+x^{10}+x^9-x^4-x^3-x^2-x-1}\\=\frac{\cancel{(x-1)}(-5x^7-10x^6-15x^5-20x^4-16x^3-12x^2-8x-4)}{\cancel{(x-1)}(x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+5x^7+5x^6+5x^5+5x^4+4x^3+3x^2+2x+1)}$$Cancel the $x-1$'s and take $x=1$. This gives: $$\frac{-90}{45}=-2$$ We chose to factor the $x-1$ since $1$ was a root of both the numerator and denominator, so $x-1$ must be a factor (and it was contributing the problematic $0$)

John Doe
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Note that $$ x^n - 1 = (x-1)(x^{n-1}+x^{n-2}+\cdots + x + 1). $$ Thus we have \begin{align} \frac{9}{x^9-1} - \frac{5}{x^5-1} &= \frac{1}{x-1}\left[\frac{9}{\sum_{i=1}^9 x^{i-1}} - \frac{5}{\sum_{i=1}^5 x^{i-1}} \right] \\ &=\frac{1}{x-1}\left[\frac{9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1}}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})}\right]. \end{align} And note that \begin{align} 9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1} &=4\sum_{i=1}^5 x^{i-1} - 5x^5\sum_{i=1}^4 x^{i-1} \\ &=4\sum_{i=1}^4 x^{i-1}(1-x^5) + x^4(4-x-x^2-x^3-x^4) \\ &=4(1-x)\left(\sum_{i=1}^4 x^{i-1}\right)\left(\sum_{i=1}^5 x^{i-1}\right) \\ &\hspace{0.5cm} +x^4(1-x)(1+(1+x)+(1+x+x^2)+(1+x+x^2+x^3)) \end{align} Therefore, \begin{align} \frac{9}{x^9-1} - \frac{5}{x^5-1} &= -\frac{4(\sum_{i=1}^4 x^{i-1})(\sum_{i=1}^5 x^{i-1}) + x^4(4+3x+2x^2+x^3)}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})} \end{align} and by letting $x \to 1$, we obtain $$ \lim_{x \to 1} \left[\frac{9}{x^9-1} - \frac{5}{x^5-1}\right] = -2. $$

induction601
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Noting $$ x^9-1=(x-1)\sum_{n=0}^8x^n, x^5-1=(x-1)\sum_{n=0}^4x^n $$ one has \begin{eqnarray} &&\lim_{x\to1}\frac{9}{x^9-1}-\frac{5}{x^5-1}\\ &=&\lim_{x\to1}\frac{9\sum_{n=0}^4x^n-5\sum_{n=0}^8x^n}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\ &=&\lim_{x\to1}\frac{9\sum_{n=0}^4(x^n-1)-5\sum_{n=0}^8(x^n-1)}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\ &=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^n(x-1)x^k-5\sum_{n=0}^8\sum_{k=0}^n(x-1)x^k}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\ &=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^nx^k-5\sum_{n=0}^8\sum_{k=0}^nx^k}{\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\ &=&\frac{9\sum_{n=0}^4(n+1)-5\sum_{n=0}^8(n+1)}{5\times9}\\ &=&-2. \end{eqnarray}

xpaul
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