Give an example to illustrate that $\lim_{x\to 0}\ f(x)$ is not always equal to $\lim_{x\to 0}\ f(2x)$

Question

It's really hard to come up with such an example. My professor said that the example won't be made up of ILATE functions. It would be something like signum, gif etc.

Answer 1

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $\Bbb{R}$, such that $\lim_{x\to 0} f(x)$ exists then $\lim_{x\to 0} f(2x)$ exists and equal to $\lim_{x\to 0} f(x)$.

I am going to prove this statement.
Let, $\lim_{x\to 0} f(x)=L\in\Bbb{R}$ and define $g:\Bbb{R}\to\Bbb{R}$ by $g(x)=f(2x)\quad\forall x\in\Bbb{R}$
To prove $\lim_{x\to 0} f(2x)=L$, it is enough to show that $\lim_{x\to 0} g(x)=L$
Choose $\varepsilon >0$
Then $\exists \delta_\varepsilon >0$ such that $|f(x)-L|<\varepsilon\quad\forall x\in(-\delta_\varepsilon,\delta_\varepsilon)\backslash\{0\} $
$\implies |f(2x)-L|<\varepsilon\quad \forall x\in\left(-\frac{\delta_\varepsilon}{2},\frac{\delta_\varepsilon}{2}\right )\backslash\lbrace 0 \rbrace$
$\implies |g(x)-L|<\varepsilon\quad\forall x\in \left (-\delta_\varepsilon',\delta_\varepsilon'\right )\backslash\{0\}$ where $\delta_\varepsilon' =\frac{\delta_\varepsilon}{2} >0$
For any $\varepsilon >0$, $\exists\delta_\varepsilon'>0$ such that $|g(x)-L|<\varepsilon\quad\forall x\in \left (-\delta_\varepsilon',\delta_\varepsilon'\right )\backslash\{0\}$ $\implies \lim_{x\to 0} g(x)=L\implies\lim_{x\to 0} f(2x)=L=\lim_{x\to 0} f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $\lim_{x\to c} f(x)$ may not be equal to $\lim_{x\to c} f(2x)$ for any $c\in\Bbb{R}$(Why?)
You can verify this result geometrically also(Hint: To obtain the graph of $x\mapsto f(2x)$, shrink the graph of $f$ horizontally by $\frac{1}{2}$).

Answer 2

By definition, if $f(\xi) \to L$ as $\xi \to 0$, this means that

$$|f(2x) - L| < \epsilon \quad \text{whenever} \quad |2x| < \delta$$

but this just means that $|x| < \delta/2$ implies $|f(2x) - L| < \epsilon$

so $f(2x) \to L$ as $x \to 0$.

Answer 3

Let $f : \{ p/3^n \mid p, n \in \mathbb N \} \to \mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $\lim_{x \to 0} f(x)$ is not defined but $\lim_{x \to 0} f(2x)$ is defined and equals $1.$