Let $f$ be a function: $$ f:\mathbb{R} \to \mathbb{R} $$ It is known that: $$ \lim_{x\to\infty}f(x) = 0 $$ I need to prove / disprove that the following limit exist: $$ \lim_{x \to 0}f\left(\frac{1}{x}\right) $$
Limits: if $ \lim_{x\to\infty}f(x) = 0 $, does $\lim_{x \to 0}f(1/x)$ exist?
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limits
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2What does $x\to \infty$ mean here? Does it mean $x \to +\infty$ or $\lvert x\rvert \to \infty$? – 2017-01-03
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1It means $$ x \to +\infty $$ – 2017-01-03
2 Answers
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Example:
$$f(x)=\begin{cases}\frac1x&;x>0\\1&;x\le0\end{cases}$$
$$\lim_{x\to\infty}f(x)=0$$
but
$$\lim_{x\to0^+}f(1/x)=\lim_{x\to0^+}x=0$$
$$\lim_{x\to0^-}f(1/x)=\lim_{x\to0^-}1=1$$
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0So simple...Thank you – 2017-01-03
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0@Noam That is my trademark ;) – 2017-01-03
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2@SimpleArt For the love of simplicity, $f(x)=e^{-x}$ is an even simpler function ;) – 2017-01-03
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1@5xum Oh you... – 2017-01-03
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Maybe the simplest counterexample would be $f(x)=e^{-x}$.
The limit $$\lim_{x\to 0} e^{-\frac1x}$$ does not exist, because if $x$ is small but positive, $e^{-\frac1x}$ is close to $0$, but if $x$ is small and negative, then $e^{-\frac1x}$ is large.