Limit (Calculus)

Question

Can you help me to solve step by step this limit, please?
(I've found out that its result is equal to 12)
$$ \lim_{n\to\infty}\frac{n\cdot \sin\left(\frac{2}{n}\right)\cdot \ln\left(1+\frac{3}{n^2}\right)} {\left(1+\frac{3}{n^2}\right)^n\cdot \left(1-\cos\left(\frac{1}{n}\right)\right)}=12 $$ Thanks in advance.

Answer 1

Hint: Try with Maclaurin's series, you should not have problems.

Answer 2

First use the substitution $n=1/x$. I think the easiest way is to use Maclaurin series: \begin{equation} \lim_{x \to 0+}\frac{\sin(2x)\ln(1+3x^2)}{x(1+3x^2)^{1/x}(1-\cos(x))} \end{equation}

\begin{equation} \lim_{x \to 0+}\frac{(2x+o(x^3))(3x^2+o(x^4))}{x(1+3x^2)^{1/x}(\frac{x^2}{2!}+o(x^4))} \end{equation}

The Maclaurin series of $g(x)=(1+3x^2)^{1/x}$ is $g(x)=g(0)+g'(0)x+\dots=1+3x+\dots$ Do yo need more help with this step?

We conclude: \begin{equation} \lim_{x \to 0+}\frac{(2x+o(x^3))(3x^2+o(x^4))}{x(1+3x+o(x^2))(\frac{x^2}{2!}+o(x^4))} \end{equation}

\begin{equation} \lim_{x \to 0+}\frac{6x^3+o(x^5)}{\frac{1}{2}x^3+\frac{3}{2}x^4+o(x^4)}=\lim_{x \to 0+}\frac{6x^3+o(x^4)}{\frac{1}{2}x^3+o(x^4)}=12 \end{equation}

Extra comments: here is the justification that the first two terms in the Maclaurin series of $g(x)=(1+3x^2)^{1/x}$ are $g(0)+g'(0)x=1+3x$. You can compute $g(0)$ by the following argument: \begin{equation} \ln g(x) = \frac{\ln(1+3x^2)}{x} \end{equation} With the l'Hôpital's rule, we find $\ln g(0) = 0$, so $g(0) = 1$. The value of $g'(0)$ is computed in the following way: \begin{equation} \ln g(x) = \frac{\ln(1+3x^2)}{x} \end{equation} After differentiation of both sides: \begin{equation} \frac{g'(x)}{g(x)} = \frac{6x^2-(1+3x^2)\ln(1+3x^2)}{x^2+3x^4} \end{equation} \begin{equation} g'(0)=\frac{g'(0)}{g(0)}=\lim_{x \to 0}\frac{g'(x)}{g(x)} = \lim_{x \to 0}\frac{6x^2-(1+3x^2)(3x^2+o(x^4))}{x^2+3x^4} = \lim_{x \to 0}\frac{3x^2+o(x^4)}{x^2+o(x^4)}=3 \end{equation}
The last limit is much faster calculated with Maclaurin series instead of applying l'Hôpital's rule!