수열의 수렴과 발산

Let's take stock: $\sin(1/n) \to \sin (0)=0$ as $n \to \infty$, so $\log\left(\sin(1/n)\right) \to -\infty$. However, $\log(2n) \to \infty$. So, we have some tension here: the two pieces behave in ways that pull the terms of the sequence in different directions. (Recall we cannot conclude anything like “$-\infty+\infty=0$.")

We try using logarithm rules to get a clearer picture.

$$\begin{aligned} \log\left(\sin\frac{1}{n}\right)+\log(2n)&=\log\left(2n\sin\left(\frac1n\right)\right) \end{aligned}$$

Still, we have indeterminate behaviour: $2n\sin(1/n)$ is the product of $2n$, which grows without bound, and $\sin(1/n)$, which approaches zero. In the past, we learned that we can handle the indeterminate form $0\cdot\infty$ with l'Hopital's rule (after a little algebra), but there's a slicker way. Note $1/n \to 0$ as $n \to \infty$. If we write $\frac{1}{n}=x$, then this piece of our limit resembles something familiar.

$$\begin{aligned} 2n\sin\left(\frac{1}{n}\right)&=2\left(\frac{\sin x}{x}\right) \\ \text{If $n \to \infty$, then $x = \frac{1}{n}\to 0$.} \\ \lim_{n \to \infty}2n\sin\left(\frac{1}{n}\right)&=2\lim_{x \to 0}\frac{\sin x}{x} \\ \text{That limit is familiar:} \\ &=2(1)=2 \\ \text{Then:} \\ \lim_{n \to \infty}\log\left(2n\sin\left(\frac{1}{n}\right)\right)&=\log 2 \end{aligned}$$

Note: if you have forgotten that $\displaystyle\lim_{x \to 0}\frac{\sin x}{x}=1$, you can also evaluate this limit using l'Hopital's rule:

$$\begin{aligned} \underbrace{\lim_{x \to 0}\frac{\sin x}{x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den}\to0}} &=\lim_{x \to 0}\frac{\cos x}{1}=\cos 0 =1 \end{aligned}$$