# Prove that $\lim_{k\rightarrow \infty}\frac{n^{2k}-1}{n^{2k-1}}=n$

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?

• Why doesn't L'Hospital rule work?Also consider that for $n=1$ this is not correct. – kingW3 Apr 9 '17 at 10:34
• Hint: $\frac{a-b}{c}=\frac ac-\frac bc$ – Hagen von Eitzen Apr 9 '17 at 10:34
• What is $n$? Is it a natural number or real? – Giulio Apr 9 '17 at 10:34
• Natural number. – MightyPython Apr 9 '17 at 10:35
• Is $n\gt1$? Because otherwise if $n=0$ we have $\frac {-1}0\neq0$ and if $n=1$ we have $\frac01=0\neq1$ – Giulio Apr 9 '17 at 10:37

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 $n>1$.
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 $n>1$
$n^{2k}-1\sim_{k\to\infty}n^{2k},\;$ hence $\enspace\dfrac{n^{2k}-1}{n^{2k-1}}\sim_{k\to\infty}\dfrac{n^{2k}}{n^{2k-1}}=n$.
• Can I write in this way: $n^{2k}-1=O(n^{2k})$ to denote that $n^{2k}\gg 1$ for large $k$? – MightyPython Apr 9 '17 at 10:49
• @MightyPython: No, because for example $2-\frac1k$ is $O(n^{2k})$ too, but is not $\gg 1$ ever. – Henning Makholm Apr 9 '17 at 10:51
• @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. – Bernard Apr 9 '17 at 10:57