Let $f:[0,\infty)\to\mathbb{R}$ be a continuous function s.t. $\forall ~x \in[0,\infty), f(x) \neq0 $ and $ \forall ~ \varepsilon>0 ~ \exists ~ x\in[0,\infty)$ $s.t. 0< f(x) <\varepsilon$. Given $\lim_{x\to\infty}f(x)=L\in\mathbb{R}$, prove $L=0$.

Hello everyone. I'm not sure what the correct approach is, but I've tried proving this by contradiction. Assume $L\neq 0 \implies $ for all $\epsilon>0$ there exists some $M\in[0,\infty)$ so that for all $x>M \rightarrow |f(x)-L|<\epsilon$.

Since $ \forall\epsilon>0.\exists x_0\in[0,\infty)$ $s.t. 0< f(x_0) <\epsilon$ and since $f$ is continuous at its domain, it follows that there exists a neighbourhood of $x_0$ such that $0<f(x)$ for all $x$ in this neighbourhood.

Let $\epsilon>0$. then there exists $M\in[0,\infty)$ s.t. for all $x>M\implies L-\epsilon<f(x)<L+\epsilon$, and there also exists $I$ s.t. for all $x>I \implies 0<f(x)<\epsilon$. Let $M_0=max\{I,M\} \implies L-\epsilon<f(x)<\epsilon$.... and then I get stuck. I would love to get your help, I want to brush up my calculus but I seem to have forgotten some epsilon-delta strategies. Thanks in advance :)


You're on the right track, but you can phrase it a lot more simply by being explicit with your choice of $\epsilon$. If $\lim_{x \to \infty} f(x) = L \ne 0$, choose $\epsilon = \frac 1 2 |L|$. Then you have

$$|f(x) - L| < \epsilon \implies |f(x)| > \frac{|L|}{2}$$

gives a contradiction.

The most important step here is to draw a picture. Draw a line at height $L \ne 0$, and put a little band around it that doesn't contain zero. The function values have to be inside this band because of the limit, and outside of the band because of the condition about $\epsilon$. This is bad.

  • $\begingroup$ Thank you very much for your answer, it helped me understand :) $\endgroup$ – Noy Perel Feb 21 at 18:54
  • $\begingroup$ I think you need to invoke the properties of continuous functions on closed intervals. The band around $L$ is to be ensured for $x>M$ whereas the values less than $\epsilon$ can happen with $x\leq M$. $\endgroup$ – Paramanand Singh Feb 21 at 18:54
  • $\begingroup$ @ParamanandSingh Yes, you're quite right. $\endgroup$ – user296602 Feb 21 at 18:59

You can note that $f$ never vanishes and takes positive values somewhere and hence is positive everywhere on $[0,\infty) $ by intermediate value theorem.

Then we have $L\geq 0$. And if $L>0$ then we can see that $f(x) > L/2$ for all $x>M$ for some $M>0$ (choose $\epsilon =L/2$ in limit definition for $\lim_{x\to \infty} f(x) =L$). Also let $m$ be the minimum value of $f$ on $[0,M]$ (remember that continuous functions take minimum and maximum values on a closed interval) and if a positive $\epsilon$ is choosen to be less than $\min(m, L/2)$ then $f(x) >\epsilon $ for all $x\in[0, \infty) $. This is contrary to the hypotheses given in question. And thus we have $L=0$.

  • $\begingroup$ Thank you very much, this is very clear and helpful! :) $\endgroup$ – Noy Perel Feb 21 at 18:53

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