$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$

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?

• On first sight, the limit seems to be infinity, using a series expansion for $\cos(\frac{1}{n^2})$. – R.J. Etienne Mar 3 at 16:09
• Thanks, but it is an exercise in a textbook in a chapter on sequences (before derivative function) then there should be a solution without series expansion. – asv Mar 3 at 16:13
• Ok, what then about using the half-angle formula $1-\cos(x)=2\sin(\frac{x}{2})$ and then $\lim\frac{\sin(x)}{x}$? – R.J. Etienne Mar 3 at 16:22

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 $$\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1$$

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 $$+\infty$$.

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

$$\begin{eqnarray*} \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))} & \stackrel{n = \frac{1}{x}}{=} & \frac{x\cdot \ln (1+x^2+x^3)}{1-\cos x^2}\\ & \stackrel{\cos x^2 > 1 - \frac{x^4}{2}}{>} & \frac{2\ln (1+x^2+x^3)}{x^3} \\ & \stackrel{L'Hosp.}{\sim} & \frac{2}{1+x^2+x^3}\left(\frac{2}{x^2} + \frac{3}{x} \right) \\ & \stackrel{x \to 0^+}{\longrightarrow} & +\infty \end{eqnarray*}$$

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 $$0$$.