Evaluate the limit of sin(x) as x approaches infinity, or show that it does not exist.

Step-by-step solution

1. 1

   Recall what the limit would have to mean. Saying $\lim_{x\to\infty}\sin x=L$ requires that for every tolerance $\varepsilon>0$ there is an $M$ beyond which $|\sin x-L|<\varepsilon$ for all $x>M$. So a single misbehaving sequence of large $x$ values is enough to destroy the claim.

2. 2

   Note that boundedness is not enough. The function satisfies $-1\le\sin x\le1$ for every $x$, so it never runs off to infinity. But being bounded does not imply convergence — a bounded function can oscillate forever, and that is exactly what happens here.

3. 3

   Exhibit two subsequences with different limits. Take

   $x_{n}=\frac{\pi}{2}+2\pi n\ \Rightarrow\ \sin x_{n}=1,\qquad y_{n}=\frac{3\pi}{2}+2\pi n\ \Rightarrow\ \sin y_{n}=-1.$

   Both $x_{n}\to\infty$ and $y_{n}\to\infty$ as $n\to\infty$, yet the function values sit permanently at $1$ and at $-1$.

4. 4

   Conclude that no limit exists. If the limit were some number $L$, then every sequence tending to infinity would have to give function values tending to $L$. Here one gives $1$ and another gives $-1$, and $1\ne-1$, so no such $L$ exists:

   $\lim_{x\to\infty}\sin x\ \text{does not exist.}$

5. 5

   Distinguish this from limits that "equal infinity". A limit can fail in two different ways: by growing without bound (as for $\lim_{x\to\infty}x^{2}=\infty$) or by oscillating (as here). Only the second applies to $\sin x$. Note also that a damped version does converge — $\lim_{x\to\infty}\frac{\sin x}{x}=0$ by the squeeze theorem, since $\left|\frac{\sin x}{x}\right|\le\frac1x$ — so it is the unbounded oscillation of the argument, not the sine itself, that causes the failure.