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Show that $\frac{5}{n}+e^{-n}=O(\frac{1}{n})$

I know that $e^n>n, (n\geq 0)$, with which $e^{-n}<\frac{1}{n}$ and so $\frac{5}{n}+e^{-n}<\frac{5}{n}+\frac{1}{n}=6\frac{1}{n}$, then $\frac{5}{n}+e^{-n}=O(\frac{1}{n})$. Is this argument right? Thank you very much.

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    $\begingroup$ It seems correct to me, since you have that $\frac{5}{n}+e^{-n}<C\frac{1}{n}$, for $n\geq n_0$, where $C=6$ and $n_0=0$. $\endgroup$ – Βασίλης Μάρκος Feb 22 '18 at 13:18
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$$\frac{\frac5n+e^{-n}}{\frac1n}=\frac{5e^n+n}{e^n}\xrightarrow[n\to\infty]{}5$$

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  • $\begingroup$ Explain a little more please, what conclusion is reached in this case? $\endgroup$ – user482152 Feb 22 '18 at 13:24
  • $\begingroup$ @user482152 That there exists a contant $\;C\;$ s.t. $$\left|\frac5x+e^{-n}\right|\le C\left|\frac1n\right|$$which is the definition of big O in your case. $\endgroup$ – DonAntonio Feb 22 '18 at 13:32

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