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.

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    $\begingroup$ How could they even be different? $\endgroup$
    – md2perpe
    Jul 18, 2018 at 12:29
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    $\begingroup$ @adityagupta, if you take the limit at $0$, they will be same. $\endgroup$
    – MathBS
    Jul 18, 2018 at 13:28
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    $\begingroup$ Either your professor tricked you, or there is something you don't tell us. $\endgroup$
    – user65203
    Jul 18, 2018 at 13:42
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    $\begingroup$ Should $f$ be defined for all $x \in \mathbb R$? If not, then I can solve it, I think. $\endgroup$
    – md2perpe
    Jul 18, 2018 at 13:43
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    $\begingroup$ This is nonsense as stated. Possibly what he asked for is not exactly what you say? $\endgroup$ Jul 18, 2018 at 13:59

3 Answers 3


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 $\frac{1}{2}$).


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 $x \to 0$.


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 $1.$

  • $\begingroup$ this was kinda the one that i thought, but yours is much more elegant. thanks for the help $\endgroup$ Jul 19, 2018 at 14:34
  • $\begingroup$ Note that one limit exists and the other does not. You can not find a function where both limits exist and are different. $\endgroup$
    – md2perpe
    Jul 19, 2018 at 19:46

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