$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$
I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?
 A: Let's write
$$\frac{\ln(1+n+n^3) - 3 \ln(n)}{n(1- \cos(1/n^2))} = \frac{\ln(1 + \frac{1}{n^2} + \frac{1}{n^3})}{n(1- \cos(1/n^2))} = \frac{\frac{1}{n^2} + \frac{1}{n^3} +o(\frac{1}{n^3})}{n (\frac{1}{2n^4} + o(\frac{1}{n^5}))} = \frac{n +1 + o(1)}{\frac{1}{2} + o(\frac{1}{n})} \sim 2n$$
so the limit is $+\infty$.
A: A long version:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}=
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{n\cdot\frac{1}{n^4}\cdot\frac{(1-\cos^2(1/n^2))}{\frac{1}{n^4}}}\cdot(1+\cos(1/n^2))=\\
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}\cdot \color{red}{\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}}}\cdot(1+\cos(1/n^2))=\\
2\cdot \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}}=
2\cdot \lim_{n \rightarrow +\infty} n^3\ln\left(\frac{1+n+n^3}{n^3}\right)=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{n^3}=\\
2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}\cdot (n+1)} = \\
2\cdot \lim_{n \rightarrow +\infty} (n+1)\cdot\ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}} =
2\cdot \ln{e} \cdot \lim_{n \rightarrow +\infty} (n+1)\rightarrow +\infty$$

On the 2nd line
$$\lim_{n \rightarrow +\infty}\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}=
\lim_{n \rightarrow +\infty}\frac{\sin(1/n^2)}{\frac{1}{n^2}}\cdot \frac{\sin(1/n^2)}{\frac{1}{n^2}}=1$$ from
$$\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1$$
A: You may proceed as follows:


*

*$\frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))} = \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}$
Now, set 


*

*$n = \frac{1}{x}$ and consider the limit for $x \to 0^+$ and use

*$\cos t > 1- \frac{t^2}{2}$ for $t \in (0,\frac{\pi}{2})$
So, you get
\begin{eqnarray*} \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}
& \stackrel{n = \frac{1}{x}}{=} & \frac{x\cdot \ln (1+x^2+x^3)}{1-\cos x^2}\\
& \stackrel{\cos x^2 > 1 - \frac{x^4}{2}}{>} & \frac{2\ln (1+x^2+x^3)}{x^3} \\
&  \stackrel{L'Hosp.}{\sim} & \frac{2}{1+x^2+x^3}\left(\frac{2}{x^2} + \frac{3}{x} \right) \\
&  \stackrel{x \to 0^+}{\longrightarrow} & +\infty
\end{eqnarray*}
A: I would be inclined to write this as $\lim_{x\to\infty}\frac{\ln(1+ x+ x^3)- 3 \ln(x)}{x(1+ \cos(1/x^2)}$ where $x$ does not have to be an integer. That way the functions are differentiable and I can use "L'Hopital's rule".
We can write the numerator as $\ln(1+ x + x^3)- 3 \ln(x)= \ln(1+ x+ x^3)- \ln(x^3)= \ln\left(\frac{1 + x+ x^3}{x^3}\right)= \ln(x^{-3} + x^{-2} + 1)$.
Its derivative is $\frac{-3x^{-4}- 2x^{-3}}{x^{-3}+ x^{-2}+ 1}= \frac{-3- 2x}{x+ x^2+ x^3}$. Since the denominator has higher degree than the numerator, the limit as $n$ (and so $x$) goes to infinity is $0$.
We can write the denominator as $x(1+ \cos(1/x))= x(1+ \cos(x^{-1}))$.
Its derivative is $1+ \cos(x^{-1})+ x(-\sin(x^{-1}))(-x^{-2})= 1+ \cos(x^{-1})- x^{-1}\sin(x^{-1})$.
As $x$ goes to infinity all those "$x^{-1}$" terms go to $0$ so the denominator goes to $1+ 1- 0= 2$.
Since that denominator is nonzero, the limit of the original expression is $0$.
