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How would one prove that:

$$ \lim_{k\rightarrow \infty}\frac{n^{2k}-1}{n^{2k-1}}=n?$$

I basically have no idea. L'Hospital seems not to work here. Any hints?

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    $\begingroup$ Why doesn't L'Hospital rule work?Also consider that for $n=1$ this is not correct. $\endgroup$ – kingW3 Apr 9 '17 at 10:34
  • $\begingroup$ Hint: $\frac{a-b}{c}=\frac ac-\frac bc$ $\endgroup$ – Hagen von Eitzen Apr 9 '17 at 10:34
  • $\begingroup$ What is $n$? Is it a natural number or real? $\endgroup$ – Giulio Apr 9 '17 at 10:34
  • $\begingroup$ Natural number. $\endgroup$ – MightyPython Apr 9 '17 at 10:35
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    $\begingroup$ Is $n\gt1$? Because otherwise if $n=0$ we have $\frac {-1}0\neq0$ and if $n=1$ we have $\frac01=0\neq1$ $\endgroup$ – Giulio Apr 9 '17 at 10:37
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Multiply numerator and denominator by $n^{-(2k-1)}$, this gives $$ \frac{n-n^{-(2k-1)}}{1} $$ whose limit is easliy seen to be $\frac n1$, assuming that $n>1$.

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Using L'Hospitals you get $$\lim_{k\to\infty}\frac{n^{2k}-1}{n^{2k-1}}=\lim_{k\to\infty}\frac{n^{2k}\log n}{n^{2k-1}\log n}=n$$ This works if $n>1$

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With equivalents:

$n^{2k}-1\sim_{k\to\infty}n^{2k},\;$ hence $\enspace\dfrac{n^{2k}-1}{n^{2k-1}}\sim_{k\to\infty}\dfrac{n^{2k}}{n^{2k-1}}=n$.

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  • $\begingroup$ Can I write in this way: $n^{2k}-1=O(n^{2k})$ to denote that $n^{2k}\gg 1$ for large $k$? $\endgroup$ – MightyPython Apr 9 '17 at 10:49
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    $\begingroup$ @MightyPython: No, because for example $2-\frac1k$ is $O(n^{2k})$ too, but is not $\gg 1$ ever. $\endgroup$ – Henning Makholm Apr 9 '17 at 10:51
  • $\begingroup$ @MightyPython: No. $n^{2k}-1=O(n^{2k})$ is true just because $n^{2k}-1\le n^{2k} $. Does it imply $n^{2k}\gg 1$? We really need equivalents to determine the limit. $\endgroup$ – Bernard Apr 9 '17 at 10:57

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