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Can somebody give me an example where $$\lim_{x\to 0}f(x^2)$$ exists but $$\lim_{x\to 0}f(x)$$ does not?

5xum
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aaaa
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  • $f(x)=\sqrt x$. – Indrayudh Roy Feb 28 '14 at 14:06
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    sign function is the best example. – Sawarnik Feb 28 '14 at 14:31
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    Many have pointed out very simple examples. In this question what one needs to see is that when we are thinking of $f(x^{2})$ we are only considering the values of $f$ for non-negative values of the argument. But in $f(x)$ we are considering both negative and positive values of the argument. Hence a counter example is created if we let $\lim_{x \to 0^{+}}f(x)$ exist and at the same time let $\lim_{x \to 0^{-}}f(x)$ not exist. All the examples given follow this. – Paramanand Singh Mar 01 '14 at 04:59
  • @ParamanandSingh : At least one example has both one-sided limits of $f$ exist. – Eric Towers Feb 03 '21 at 17:01
  • @EricTowers: you are right! But I can't edit my comment now. – Paramanand Singh Feb 04 '21 at 00:43

4 Answers4

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Yes. Let $f(x) = 1$ for $x<0$ and $f(x) = 0$ for $x \geq 0$. $f(x^2) = 0$ so the limit exists. $f(x)$ is has a jumpt discontinuity at zero (the limit from the left is $1$ but the limit from the right is zero) so the limit does not exist.

Eric Towers
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For example

$$f(x):=\frac{|x|}x\implies\;\lim_{x\to 0}f(x)=\;\text{does not exist (check one sided limits), but}$$

$$\lim_{x\to 0}f(x^2)=\lim_{x\to 0}\frac{x^2}{x^2}=1\;\;\text{does exist}\;\ldots$$

The first part above is nothing else but the proof of the well known fact that the absolute value function isn't differentiable at zero.

learner
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DonAntonio
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    This is hardly any different from what what posted and upvoted 6hs ago. – Pedro Feb 28 '14 at 14:33
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    Didn't even see it, @PedroTamaroff, yet after checking I trust you're able to see it is way a different example of a function than the one given there, and in this case it is a rather familiar function... – DonAntonio Feb 28 '14 at 14:37
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Consider $f(x) = \chi_{[0,\infty)}(x)$, the characteristic function of the interval $[0,\infty)$.

J. J.
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$\lim\limits_{x \to 0} \sqrt{x}$ would be a good choice because the domain of f(x) is not a punctured neighbourhood of zero. Whereas it is trivial that $\lim\limits_{x \to 0} {x}$$ exists

HHHHH
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