How can I test the condition and stability of the following function, for values $x=0$, $x=0.25$ and $x=10^{-5}$? $$ f(x) = \frac{1-cos(2x)}x $$ What should one do here? For condition, I think I should do the following: $$ K = \left|\frac{f'(x)}{f(x)} \cdot x\right| $$ and check if it's $>> 1$ or $< 10$.

1) Am I right for the condition? And how to check for that $x$ values given above? If not, how would be the right way?

2) How to check the stability also for that values?

Thank you in advance!!

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    $\begingroup$ You could say $\cos 2x = 1-2\sin^2 x$ and then $f(x) = \frac {2\sin^2 x}{x}$ and you might have some experience already with the function $\frac {\sin x}{x}$ $\endgroup$ – Doug M Jul 19 '18 at 21:34
  • $\begingroup$ because of the undefined form 0/0 you mean? $\endgroup$ – ZelelB Jul 19 '18 at 21:44
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    $\begingroup$ Your function is undefined at $0.$ However, the limit exists. $\endgroup$ – Doug M Jul 19 '18 at 21:45
  • $\begingroup$ yep, but how to answer the questions above? What should I do in the exam for this problem? $\endgroup$ – ZelelB Jul 19 '18 at 21:45
  • $\begingroup$ would be greateful for any help! I have my last exam tomorrow, and this comes every time :/ $\endgroup$ – ZelelB Jul 19 '18 at 22:19

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