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We have $|x_2-x_1|<\delta \implies |f(x_1)-f(x_2)|<\varepsilon$ so $|\cos x_1-\cos x_2|<\varepsilon$ how should I proceed after this? There is already an answer using Lipschitz and Mean value theorem but is there any way to do this question by not using them? Because in the book which I am using these topics are given after this question.

Theo Bendit
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hari haran
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$cos(x)$ is continuous and is therefore uniformly continuous on compact intervals $[a,b]$. Therefore, it is uniformly continuous on $[0,2\pi]$. By periodicity, we get uniform continuity on $\mathbb{R}$.

Eric
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  • Hi can you solve using the method I mentioned? – hari haran Sep 04 '23 at 17:40
  • @hariharan We are not doing your home work. – Kurt G. Sep 04 '23 at 17:48
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    @KurtG. I dont want you to do my homework , like I said above I am stuck in THAT step and I want you guys to help in that. – hari haran Sep 04 '23 at 17:50
  • @hariharan For an $~\epsilon, ~\delta~$ approach, note that if $~f(x) = \cos(x),~$ then $~|f'(x)| = |\sin(x)| \in [0,1].~$ Therefore, if $~0 < |x - a| < \delta,~$ then you should be able to show that $~|f(x) - f(a)| < \delta.~$ You mentioned that you wish to avoid the Mean Value Theorem. Okay, look at the proof to the Mean Value Theorem, and (in effect) re-invent the wheel, incorporating the proof directly into your analysis. The alternative strategy to proving the key point in the $~\epsilon, ~\delta~$ approach is to use the (easily proven) identity shown in Theo Bendit's comment. – user2661923 Sep 04 '23 at 18:08
  • @user2661923 Thank you – hari haran Sep 04 '23 at 18:11