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, Henrik Aug 12 '16 at 15:20

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  • $\begingroup$ Have you tried applying the definitions of "strictly decreasing" or "$\lim_{x\to\infty}f(x)=l$"? What did that get you? $\endgroup$ – Cameron Buie Aug 12 '16 at 11:43
  • $\begingroup$ 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 $|f(x)-l|<\epsilon$ . $\endgroup$ – user90533 Aug 12 '16 at 11:51
  • $\begingroup$ $l-\epsilon <f(x)<l+\epsilon $, but from this how to get the required thing $\endgroup$ – user90533 Aug 12 '16 at 11:55
  • $\begingroup$ Taking $\epsilon =0$ we get $l<f(x)$ but where do we used decreasing criteria? $\endgroup$ – 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?

  • $\begingroup$ Can u elaborate this... $\endgroup$ – user90533 Aug 12 '16 at 12:01
  • 1
    $\begingroup$ @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. $\endgroup$ – Vincenzo Oliva Aug 12 '16 at 12:06

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