# Limit at infinity for decreasing function [closed]

Suppose $f$ is strictly decreasing function on $(0, \infty)$and $\lim_{x\rightarrow \infty} f(x)=l$ then prove that $l<f(x)$ for all $x \in (0, \infty$).

## closed as off-topic by Did, R_D, ervx, naslundx, HenrikAug 12 '16 at 15:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, R_D, ervx, naslundx, Henrik
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you tried applying the definitions of "strictly decreasing" or "$\lim_{x\to\infty}f(x)=l$"? What did that get you? – Cameron Buie Aug 12 '16 at 11:43
• I tried by definition $\lim_{x\rightarrow \infty} f(x)=l$ I.e., for every $\epsilon >0$, there exists $k$ depending $\epsilon$ snd for any $x>k$ such that $|f(x)-l|<\epsilon$ . – user90533 Aug 12 '16 at 11:51
• $l-\epsilon <f(x)<l+\epsilon$, but from this how to get the required thing – user90533 Aug 12 '16 at 11:55
• Taking $\epsilon =0$ we get $l<f(x)$ but where do we used decreasing criteria? – user90533 Aug 12 '16 at 11:57

Hint: What can you say about all the numbers in $(x_0,+\infty)$, if $f(x_0)<l$ for some $x_0\in(0,+\infty)$? What can you say about $\lim\limits_{x\to+\infty}f(x),$ then?
• @user90533: The idea is to prove the statement by contradiction. For this purpose, assume $f(x_0)<l.$ What does the decreasingness of $f$ tell you about $f(x)$, with $x>x_0$? In the end, you should be able to compare $f(x)$ ($x>x_0$) and $l$ and get a contradiction. – Vincenzo Oliva Aug 12 '16 at 12:06