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let $f$ be a continuous function defined in $\mathbb{R}$.

If

$$ \lim_{x\rightarrow + \infty} f(x) = l > 0 $$

can I say that there's an $\tilde{x} \in \mathbb{R}$ such that $f(\tilde{x}) > 0$? My guess is yes, because by the definition of limit

$$ \forall \epsilon>0 \exists x_0 : \forall x > x_0 \; |f(x) - l|<\epsilon $$ specifically $$ \forall \epsilon>0 \exists x_0 : \forall x > x_0 \; -\epsilon < f(x) - l <\epsilon $$

For $\epsilon = l$ I specifically have

$$ \exists x_0 : \forall x > x_0 -l < f(x) - l \Rightarrow \exists x_0 : \forall x > x_0 \; f(x) > 0 $$

Which means I can pick an $\tilde{x} > 0$ such that $f(\tilde{x}) > 0$.

Two further questions If this is true I assume that if

$$ \lim_{x\rightarrow - \infty} f(x) = l < 0 $$

I can say that there's an $\tilde{x} \in \mathbb{R}$ such that $f(\tilde{x}) < 0$, is that right?

Combining both

\begin{equation} \begin{array}{l} \lim_{x\rightarrow - \infty} f(x) = l_{-} < 0 \\ \lim_{x\rightarrow + \infty} f(x) = l_{+} > 0 \end{array} \end{equation}

I can then say there's an interval $A = [x_{-},x_{+}]$ such that for some $\tilde{x} \in A$ we have $f(\tilde{x}) = 0$.

Is that right?

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    It's all quite right! Well done! – 2017-01-13
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    Is this a well know result that I'm not aware of? – 2017-01-13
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    I'd say that the two steps 1) if $f$ has both a positive and negative limit on $\mathbb{R}$, then it has both positive and negative values (by the definition of limit) and 2) if $f$ is continuous on $\mathbb{R}$ and assumes both positive and negative values then it's $0$ somewhere (by the intermediate value theorem/ Bolzano's theorem) are so commonly used that ... it's not quite a well known result, but it's a well recognized result :) Bonus: a) Which other values can you prove that $f$ assumes under your conditions? b) Can you find an $f$ that assumes *only* the values you proved in a)? – 2017-01-13
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    a) Any value in $(l_{-},l_{+})$ by the intermediate value theorem b) $f(x) = l_{-} + (l_{+}-l{-})\frac{1}{1+e^{-x}}$, are those right? – 2017-01-13
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    Yes, well done! – 2017-01-13

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