2

My Maple input limit(sin(1/n)*n,n=infinity); says 1.

I don't understand why $$ \lim_{n \to \infty} \sin\left(\frac{1}{n}\right) \cdot n = 1 $$

I know that $\lim_{n \to \infty} 1/n = 0$, so it kind of says "0 * infinity = 1".

Have I overlooked some rewriting of $\sin(1/n) n$?

Jamgreen
  • 809

4 Answers4

7

Make a change of variables $x={1\over n}$ so that $x\to 0$ as $n\to \infty$, then use $\lim_{x\to 0} {\sin x\over x}=1$.

Matt Dickau
  • 2,227
  • It should be noted that $\lim_{x \to 0} \frac{\sin{x}}{x} = 1$ is a standard limit and is proved here. –  Mar 12 '19 at 12:52
3

$$n\sin\left(\frac{1}{n}\right)=\frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}$$

MSE
  • 3,153
2

$$\lim_{n \to \infty} \sin\left(\frac{1}{n}\right) \cdot n = \lim_{n \to 0} \frac{\sin n}{n}=1$$

E.H.E
  • 23,280
1

Because $$\lim_{x\to 0}\frac{\sin x}{x} = 1,$$which can most easily seen by the Taylor expansion of $\sin x$ about $0,$ $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \dots.$$ Taking $x$ as $1/n$ reveals the deep equivalence.

CR Drost
  • 2,912