Example 3
Show that \(\lim\limits_{x \rightarrow 0} \sin\left(\frac{1}{x}\right)\) doesn't exist.

Proof

Let \(f(x) = \sin\left(\frac{1}{x}\right)\). Consider the two subsequences

$$ \begin{align*} x_n = \frac{1}{2n\pi + \frac{\pi}{2}} \quad \text{ and } \quad y_n = \frac{1}{2n\pi - \frac{\pi}{2}} \end{align*} $$

Then clearly, \(\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n = 0\). Moreover, observe that

$$ \begin{align*} \sin\left(\frac{1}{x_n}\right) = \sin\left( 2n\pi + \frac{\pi}{2} \right) = 1 \quad \text{ and } \quad \sin\left(\frac{1}{y_n}\right) = \sin\left( 2n\pi - \frac{\pi}{2} \right) = -1 \end{align*} $$

Hence,

$$ \begin{align*} \lim\limits_{n\to \infty} f(x_n) = 1 \quad \text{ and } \quad \lim\limits_{n\to \infty} f(y_n) = -1 \end{align*} $$

Therefore, by the Sequential Criterion Theorem, \(\lim\limits_{x\to 0} \sin\left(\frac{1}{x}\right)\) doesn’t exist. \(\blacksquare\)

References