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$$f(x) = \begin{cases}(1-\cos x)/x & x \neq 0\\0& x=0\end{cases}$$

I am asked to prove if it is continuous at $x_1=0$

$$|f(x)−f(c)|<\varepsilon$$

Since $$1-\cos(x)=2\sin^2(x/2)$$

then $$\left|\frac{2\sin^2(x/2)}{x}-0\right|<\varepsilon$$

I let $|x-c|<\delta$, thus $|x|<\delta$. By the Archimedean Principle, $1/|x| < \delta$

Since $\sin^2(x/2)$ lies between 0 and 1, then $$\left|\frac{2}{x}\right|<\varepsilon$$

Now let $\delta=\varepsilon/2$, then $$\left|\frac{2}{x}\right|<2\left(\frac{\epsilon}{2}\right)=\epsilon$$

So it is continuous?

I tried as best as I could to code. If anyone could point out where my mistake is in proving, I would appreciate it as I'm struggling with real analysis.

rubik
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Faithhhhhh
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2 Answers2

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Unless you are really requested to prove continuity by $\epsilon$ and $\delta$, why don't you simply remark that $$ \begin{align} \lim_{x \to 0} \frac{1-\cos x}{x} &= \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)} = \lim_{x \to 0} \frac{\sin^2 x}{x} \frac{1}{1+\cos x} \\ &= \lim_{x \to 0} x\frac{\sin^2 x}{x^2} \frac{1}{1+\cos x} = 0 \cdot 1 \cdot \frac12 =0? \end{align} $$

Siminore
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  • thank you i didnt know i could do it this way ! the question was to prove whether it is continuous or not! however if i did it by the epsilon delta method, is it correct in this case? – Faithhhhhh Nov 20 '14 at 09:41
  • @Siminore, I don't see how the first equality follows. Shouldn't that be as follows?

    $$\lim_{x \rightarrow 0}{\frac{1 - \cos(x)}{x}} = \lim_{x \rightarrow 0}{\frac{(1 - \cos(x))(1 + \cos(x))}{x(1 + \cos(x))}}$$

     – Jose Arnaldo Bebita Dris Nov 20 '14 at 12:03
  • Sorry, my fault: I wrote sin instead of cos! – Siminore Nov 20 '14 at 12:06
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You could try using L'Hopital's Rule to show that:

$$\lim_{x \rightarrow 0}{\frac{1 - \cos(x)}{x}} = \lim_{x \rightarrow 0}{\frac{\sin(x)}{1}} = 0.$$

Jose Arnaldo Bebita Dris
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