I want to solve this $$ \lim_{n \to \infty} \frac{\log n^3}{\log \left(n^3+3n^2\right)}$$ I found that $$ \frac{\log n^3}{\log \left(n^3+3n^2\right)}<\frac{\log\left( n^3\right)}{\log \left(n^3\right)}=1$$ but I need another bond according to the squeeze theorem.
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$$=\lim_{n\to\infty}\dfrac{3\ln n}{3\ln n+\ln\left(1+\dfrac3n\right)}=\lim_{n\to\infty}\dfrac{3}{3+\dfrac{\ln\left(1+\dfrac3n\right)}{\ln n}}=?$$
lab bhattacharjee
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bhattacharje really I can't understand – Anne Nov 28 '17 at 13:47
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@Anne, $$\ln(n^a)=a\ln n$$ right? – lab bhattacharjee Nov 28 '17 at 13:48
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I can't understand the chamgements in the denominator – Anne Nov 28 '17 at 13:49
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@Anne, Divided numerator & denominator by $\ln n$ – lab bhattacharjee Nov 28 '17 at 13:50
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ahh ok now I see...thanks for your help – Anne Nov 28 '17 at 13:51
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If you use equivalents, it's obvious: $\;n^3+3n^2\sim_\infty n^3$, so $$\log(n^3+3n^2)\sim_\infty\log n^3,\quad\text{and finally }\;\frac{\log n^3}{\log(n^3+3n^2)}\sim_\infty\frac{\log n^3}{\log n^3}=1.$$
Bernard
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