The proof of 'the limit of $\sin(1/x)$ as $x$ approaches $0$ does not exist'

Question

How can I prove that the limit of $\sin(1/x)$ as $x$ approaches $0$ does not exist? Should I use $\varepsilon$-$\delta$ in order to prove it? Are there any alternative ways to prove it?

Answer 1

If $x = 1/(\pi(2n+1/2)) $ then $\sin(1/x) =\sin(\pi(2n+1/2)) = 1 $.

If $x = 1/(\pi(2n+3/2)) $ then $\sin(1/x) =\sin(\pi(2n+3/2)) = -1 $.

Therefore $\lim_{x \to 0} \sin(1/x) $ does not exist.

Just choose $\epsilon = 1$ in a standard proof.

Answer 2

There are many ways you could potentially prove this, but one simple way to do this by first principles (i.e., without invoking any other limit rules) is to prove directly that the epsilon-delta condition for the limit cannot hold. To do this, take $n \in \mathbb{N}$ and consider the two points:

$$x_* = \frac{2}{\pi (2n+1)} \quad \quad \quad x_{**} = -\frac{2}{\pi (2n+1)}.$$

These points give $\sin(1/x_*) = (-1)^n$ and $\sin(1/x_{**}) = (-1)^{n+1}$ so we have:

$$\Big| \sin \Big( \frac{1}{x_*} \Big) - \sin \Big( \frac{1}{x_{**}} \Big) \Big| = 2.$$

For any asserted limit value $L$, we can use the triangle inequality to obtain:

$$\Big| \sin \Big( \frac{1}{x_*} \Big) - L \Big| + \Big| \sin \Big( \frac{1}{x_{**}} \Big) - L \Big| \geqslant \Big| \sin \Big( \frac{1}{x_*} \Big) - \sin \Big( \frac{1}{x_{**}} \Big) \Big| = 2.$$

It follows that:

$$\max \Bigg( \Big| \sin \Big( \frac{1}{x_*} \Big) - L \Big|, \Big| \sin \Big( \frac{1}{x_{**}} \Big) - L \Big| \Bigg) \geqslant 1.$$

This is now enough for us to show that the epsilon-delta condition for the limit cannot hold. For any value $\delta > 0$ we can choose $n$ to be large enough that $0 < |x_*| = |x_{**}| < \delta$, which means that both points fall within the domain for the antecedent condition for the epsilon-delta definition of the limit. Taking $\epsilon < 1$ we see that it is impossible for the consequent condition to hold for both $x_*$ and $x_{**}$, so the delta-epsilon condition is not satisfied.

Answer 3

Let $\displaystyle x_n = \frac{1}{n\pi}$ and $y_n = \dfrac{2}{(4n+1)\pi}$ then

$\displaystyle \sin\left( \frac{1}{x_n}\right) = \sin(n \pi) = 0$ and $\displaystyle \sin\left( \frac{1}{y_n} \right) = \sin\left( \frac{(4n+1)\pi}{2}\right) = 1$

Note that $x_n \to 0$ and $y_n \to 0$ but $\displaystyle \sin\left( \frac{1}{x_n}\right) \to 0$ and $\displaystyle \sin\left( \frac{1}{y_n} \right) \to 1$ respectively. Thus, there is no limit.