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 🙂
Best Answer
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.