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Which function grows faster

$𝑓(𝑛)= 2^{𝑛^2+3𝑛}$ and $𝑔(𝑛) = 2^{𝑛+1}$

by using the limit theorem I will first simplify

then I will just get $$\lim_{n \to \infty} \dfrac{2^{n^2+3n}}{2^{n+1}}=\lim_{n \to \infty} 2^{n^2+3n-n-1}=\lim_{n \to \infty} 2^{n^2+2n-1}=\infty$$

Is this enough? I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!

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  • $\begingroup$ I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents. $\endgroup$ – lulu Oct 26 '18 at 14:49
  • $\begingroup$ @lulu could you see the question again I edit the 1 , must be on the power of 2 $\endgroup$ – NANA Oct 26 '18 at 14:57
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    $\begingroup$ Yes. With that edit , your arithmetic is correct. And your argument is sufficient. $\endgroup$ – lulu Oct 26 '18 at 15:07
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Before Edit: Your idea was correct, but you didn’t simplify the limit properly. $$\lim_{n \to \infty} \frac{2^{n^2+3n}}{2^n+1}$$ It is enough to divide both the numerator and denominator by $2^n$. $$\lim_{n \to \infty} \frac{\frac{2^{n^2+3n}}{2^n}}{\frac{2^n+1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+3n-n}}{2^{n-n}+\frac{1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+2n}}{1+\frac{1}{2^n}}$$ As $n \to \infty$, it becomes clear that the limit tends to $\infty$ since the numerator tends to $\infty$ while the denominator tends to $1$.

After Edit: Yes, your way is correct.

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  • $\begingroup$ I forget to add brackets on the second function, it must be $2^{n+1}$ @KM101 $\endgroup$ – NANA Oct 26 '18 at 14:56
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    $\begingroup$ Well, I guess all the answers are pointless now. :-) $\endgroup$ – KM101 Oct 26 '18 at 15:04
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    $\begingroup$ sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101 $\endgroup$ – NANA Oct 26 '18 at 15:06
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It is $$\frac{2^{n^2}\cdot 2^{3n}}{2^n\left(1+\frac{1}{2^n}\right)}=\frac{2^{n^2+2n}}{1+\frac{1}{2^n}}$$

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HINT

You conclusion is correct but that step is wrong

$$\lim_{n \to \infty}= \dfrac{2^{n^2+3n}}{2^n+1}\color{red}{=\lim_{n \to \infty} 2^{n^2+3n-n-1}}$$

you could use that $2^n+1\le 2^{n+1}$ and therefore

$$\dfrac{2^{n^2+3n}}{2^n+1}\ge \dfrac{2^{n^2+3n}}{2^{n+1}}$$

Update after editing

For $g(n)=2^{n+1}$ your method is absolutely correct.

What about $f(n)=2^{n^2+3n}$ and $g(n)=3^{n+1}$?

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  • $\begingroup$ I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi $\endgroup$ – NANA Oct 26 '18 at 15:17
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    $\begingroup$ @NANA We could use $3^{n+1}\le 4^{n+1}$. $\endgroup$ – gimusi Oct 26 '18 at 15:34
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    $\begingroup$ $3^{n+1}=2^{(log_23)(n+1)}$ $\endgroup$ – TurlocTheRed Oct 26 '18 at 17:35

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