Convergence of the sequence $2n\times\sin(\frac{1}{n-1})$? I have the sequence $$\{2n*\sin(\frac{1}{n-1})\}$$ and I'm supposed to see if this sequence converges. By transforming this sequence into a function of $f(x)$ and applying L'Hopital's rule to the limit, I get something like$$\lim_{x\to\infty}\frac{2{\frac{-1}{(x-1)^2}}\cos\frac{1}{x-1}}{\frac{-1}{x^2}}$$
Now I simplify the expression by dividing $\frac{-1}{(x-1)^2}$ with $\frac{-1}{x^2}$ and ultimately get that the limit equals 2.
Is this correct? Am I allowed to simplify like that? I am guessing yes because they are both polynomials of the same degree and if the degrees of the first terms match then we can safely take their quotient as 1.
 A: Yes, your answer is correct. Using L'Hôpital's rule,
$$
\lim_{x\to\infty}\Bigl[2x\sin\Bigl(\frac1{x-1}\Bigr)\Bigr]=\lim_{x\to\infty}\Bigl[2\frac{x^2}{(x-1)^2}\cos\Bigl(\frac1{x-1}\Bigr)\Bigr]=2
$$
using the fact that $x^2/(x-1)^2\to1$ and $\cos(1/(x-1))\to1$ as $x\to\infty$.
Alternatively, we can use the fact that $\sin(x)/x\to1$ as $x\to0$. We have that
$$
2n\sin\Bigl(\frac1 {n-1}\Bigr)=2\frac{n}{n-1}\frac{\sin\bigl(\frac1 {n-1}\bigr)}{\frac1{n-1}}\to2
$$
as $n\to\infty$ since $n/(n-1)\to1$ and $(n-1)\sin(1/(n-1))\to1$ as $n\to\infty$.
A: Close to $0^+$, one can bound the sine by $x-x^3/6 \le \sin(x) \le x$, which allows you to lower and upper bound your series, and get the result.
The use of the de l'Hôpital rule should be handled with care, and you are advised to state the assumptions correctly. If I understand your method correctly, in $]0,1[$ the functions $f(x) = \sin\left(\frac{1}{x-1} \right)$ and $g(x) =\frac{1}{x} $ are differentiable, and the limit $f'/g'$ exists. In its  standard form, it can be applied when $f(x)$ and $g(x)$ tends to $0$ or $\pm \infty$, which is  the case here. 
