got this questions from homework and don't have a clue how to start. I know that every real function which is continuous at a closed interval, is considered to be uniform continuous at this interval. However, my problem is how to deal with interval like this: $[a,\infty)$ , $(a,b)$ . I need to prove that $\frac{1-sinx}{cosx} $ is uniform continuous at $(0,0.5\pi)$. And also the function ${\sqrt{x}}sin \frac {1}x{}$ is uniform continuous at the interval $(0,\infty)$. Any clues, direction or ways will be helpful. Thank you in advance.
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Let $f : (a, b) → \mathbb{R}$ continuous. $f$ is uniformly continuous if and only if $\lim_{x \to a} f(x)$ and $\lim_{x \to b} f(x)$ exist .
$f(x)=\frac{1- \sin x}{\cos x}$ is continuous.
$$\lim_{x \to 0} f(x)=1$$
$$\lim_{x \to \frac{\pi}{2}} f(x)=\lim_{x \to \frac{\pi}{2}} \frac{1-\sin x}{\cos x}=\lim_{x \to \frac{\pi}{2}} \frac{- \cos x}{- \sin x}=\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\sin x}=0$$
So $f$ is uniformly continuous.
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1how did you come to this function? which identities you used? – user3708158 Aug 30 '14 at 17:46
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Do you mean at the point $$\lim_{x \to \frac{\pi}{2}} \frac{1-\sin x}{\cos x}=\lim_{x \to \frac{\pi}{2}} \frac{- \cos x}{- \sin x}$$ ? At this equality I used the De L'Hospital Rule. Are you familiar with that? – Aug 30 '14 at 23:38