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.
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1$\begingroup$ How did you find that the value of the limit is $12$? $\endgroup$– Pierre-Guy PlamondonMay 6, 2015 at 13:20
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$\begingroup$ Thanks to Wolfram Alpha, to be honest. $\endgroup$– QbitMay 6, 2015 at 13:22
2 Answers
Hint: Try with Maclaurin's series, you should not have problems.
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!
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$\begingroup$ I've really appreciated your help, it's far clear now! Thank you so much! $\endgroup$– QbitMay 6, 2015 at 14:23
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$\begingroup$ @Qbit: if you want to thank me, please rate it. Thx in advance! $\endgroup$ May 6, 2015 at 14:36