Proving Limit of Piecewise Function at x=0

[Harambe1]

Hi,

I'm struggling to prove that a limit ceases to exist as x tends to 0 for the following function:

$$f(x)=\begin{cases}\sin(\frac{1}{x}), & \text{if $x \notin \mathbb{Q}$} \\[3pt] 1, & \text{if $x \in \mathbb{Q}$} \\ \end{cases}$$

I've attempted a proof by contradiction, assuming the limit is $L$ but I'm not sure how to then use the epsilon-delta definition to prove that a limit doesn't exist.

Any tips appreciated. Thanks.

 

[Euge]

Welcome, Harambe! (Wave)

Let's consider what it means for $\lim_{x\to 0} f(x) = L$. By definition, it means that to every $\epsilon > 0$ there corresponds $\delta > 0$ such that for all $x\in \Bbb R$, $0 < \lvert x\rvert < \delta$ implies $\lvert f(x) - L\rvert < \epsilon$. Let's negate this: $\lim_{x\to 0} f(x) \neq L$ if there exists $\epsilon > 0$ such that for every $\delta > 0$, there is an $x\in \Bbb R$ such that $0 < \lvert x \rvert < \delta$ and $\lvert f(x) - L\rvert \ge \epsilon$. Now we have an $\epsilon-\delta$ formulation for the statement that $f(x)$ has no limit as $x \to 0$:

To every $L \in \Bbb R$, there corresponds an $\epsilon > 0$ such that for every $\delta > 0$, there exists an $x\in \Bbb R$ such that $0 < \lvert x\rvert < \delta$ and $\lvert f(x) - L\rvert \ge \epsilon$.

To prove the above statement, let $L\in \Bbb R$ and consider separate cases (a) $\lvert L \rvert \neq 1$ and (b) $\lvert L\rvert = 1$.

 

[Opalg]

Another way to tackle this problem would be to use sequences. Suppose that $\lim_{x\to0}f(x) = L.$ It follows that if $\{x_n\}$ is a sequence with $\lim_{n\to\infty}x_n = 0$ then $\lim_{n\to\infty}f(x_n) = L.$

For your function $f(x)$, if $\{x_n\}$ is a sequence of rational numbers tending to $0$, then $f(x_n) = 1$ for all $n$. It follows that if $\lim_{x\to0}f(x)$ exists then that limit must be $1$. If you can then find a sequence of irrational numbers $\{x_n\}$ such that $\lim_{x\to0}f(x_n)$ is some number different from $1$, it would follow that $\lim_{x\to0}f(x)$ does not exist.