Limits of sin(1/x) at 0 and at infinity
Given the function $f(x) = \sin\left(\dfrac{1}{x}\right)$, find $\displaystyle\lim_{x\to+\infty} f(x)$, and show that $\displaystyle\lim_{x\to0^+} f(x)$ and $\displaystyle\lim_{x\to0^-} f(x)$ do not exist.
Explanation of Method
Everything here hinges on the behavior of the inner quantity $\frac1x$. When $x\to+\infty$, $\frac1x\to0$, a single finite value, so continuity of $\sin$ lets us evaluate the limit by direct substitution. But when $x\to0^{\pm}$, $\frac1x\to\pm\infty$, and $\sin$ of an argument tending to infinity does not settle down — it keeps oscillating between $-1$ and $1$ forever, which can be shown rigorously by exhibiting sequences along which $\sin(1/x)$ takes different values.
- Evaluate $\lim_{x\to+\infty}\sin(1/x)$ using continuity of $\sin$ at $0$.
- Exhibit sequences $x_n\to0^+$ along which $\sin(1/x_n)$ takes different values, to show the limit does not exist.
- Repeat the same argument for $x\to0^-$.
Step 1 · Limit as $x \to +\infty$
As $x \to +\infty$, $\dfrac1x \to 0$. Since $\sin$ is continuous everywhere, in particular at $0$, the limit of the composition equals $\sin$ evaluated at the limit of the inside:
$$ \lim_{x \to +\infty} \sin\left(\frac1x\right) = \sin\left(\lim_{x\to+\infty}\frac1x\right) = \sin(0) = 0. $$
Step 2 · Show $\lim_{x\to0^+}\sin(1/x)$ does not exist
Consider the sequence $x_n = \dfrac{1}{\frac{\pi}{2}+2n\pi}$ for $n=0,1,2,\dots$, which tends to $0^+$ as $n\to\infty$. Along this sequence, $\dfrac{1}{x_n} = \dfrac{\pi}{2}+2n\pi$, so:
$$ \sin\left(\frac{1}{x_n}\right) = \sin\left(\frac{\pi}{2}+2n\pi\right) = 1 \quad \text{for every } n. $$
Now consider the sequence $y_n = \dfrac{1}{\frac{3\pi}{2}+2n\pi}$, which also tends to $0^+$. Along this sequence:
$$ \sin\left(\frac{1}{y_n}\right) = \sin\left(\frac{3\pi}{2}+2n\pi\right) = -1 \quad \text{for every } n. $$
Two sequences both tending to $0^+$ give $\sin(1/x)$ tending to two different values ($1$ and $-1$), so the limit cannot exist:
$$ \lim_{x \to 0^+} \sin\left(\frac1x\right) \ \text{does not exist.} $$
Step 3 · Show $\lim_{x\to0^-}\sin(1/x)$ does not exist
The identical argument applies with negative sequences, e.g. $x_n = \dfrac{1}{-\frac{\pi}{2}-2n\pi} \to 0^-$ gives $\sin(1/x_n) = \sin\left(-\frac{\pi}{2}-2n\pi\right) = -1$ for every $n$, while $y_n = \dfrac{1}{-\frac{3\pi}{2}-2n\pi}\to0^-$ gives $\sin(1/y_n) = \sin\left(-\frac{3\pi}{2}-2n\pi\right) = 1$ for every $n$. Again two different limiting values along sequences tending to the same point, so:
$$ \lim_{x \to 0^-} \sin\left(\frac1x\right) \ \text{does not exist.} $$
Summary Table
| Limit point | Behavior of $\frac1x$ | Behavior of $\sin(1/x)$ |
|---|---|---|
| $x\to+\infty$ | $\to 0$ (single value) | $\to \sin(0)=0$ |
| $x\to0^+$ | $\to+\infty$ | oscillates forever — DNE |
| $x\to0^-$ | $\to-\infty$ | oscillates forever — DNE |
Additional Notes
- This is one of the most classic examples in real analysis of a bounded function with no limit at a point: $\sin(1/x)$ is trapped between $-1$ and $1$ everywhere, yet still fails to converge as $x\to0$ from either side, because it oscillates infinitely often in any neighborhood of $0$.
- The contrast with $x\to+\infty$ is instructive: there, the inner function $\frac1x$ approaches a single finite value ($0$), so continuity of $\sin$ applies directly; near $x=0$, the inner function diverges to $\pm\infty$, and no amount of continuity of $\sin$ can rescue a limit that doesn't exist for its argument.
- Multiplying $\sin(1/x)$ by a factor that vanishes at $0$, such as $x$ (giving $x\sin(1/x)$), tames this oscillation via the Squeeze Theorem and produces a limit of $0$ at $x=0$ — a good related exercise.
Result
The three limits of $f(x)=\sin(1/x)$ are: