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So, here's what I'm trying to prove:

Let $x_0 \in \mathbb{R}$. Then, $\lim_{x \to x_0} \cos(x) = \cos(x_0)$

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Proof Attempt:

Let $\epsilon > 0$. Then, we have to show that:

$$\exists \delta > 0: 0 < |x-x_0| < \delta \implies |\cos(x)-\cos(x_0)| < \epsilon$$

I'm going to start by simplifying that expression with the cosines:

$$\cos(x) = \cos[\frac{x+x_0}{2}+\frac{x-x_0}{2}]$$

$$\cos(x) = \cos(\frac{x+x_0}{2})\cos(\frac{x-x_0}{2}) - \sin(\frac{x+x_0}{2})\sin(\frac{x-x_0}{2})$$

$$\cos(x_0) = \cos[\frac{x_0+x}{2}+\frac{x_0-x}{2}]$$

$$\cos(x_0) = \cos(\frac{x_0+x}{2})\cos(\frac{x_0-x}{2}) - \sin(\frac{x_0+x}{2})\sin(\frac{x_0-x}{2})$$

$$\cos(x)-\cos(x_0) = 2\sin(\frac{x_0+x}{2})\sin(\frac{x_0-x}{2})$$

So, we have:

$$|\cos(x)-\cos(x_0)| = 2|\sin(\frac{x_0+x}{2})| \cdot |\sin(\frac{x_0-x}{2})| \leq 2 |\sin(\frac{x_0-x}{2}|$$

$$|\cos(x)-\cos(x_0)| \leq 2 \cdot |\frac{x_0-x}{2}| = |x-x_0| < \delta$$

So, let $\delta = \epsilon$ and that proves the given assertion.

Could someone check through the proof above and see if I'm correct or not? I'm sorry if anything seems out of place, I'll try to fix it if the errors are just too large to ignore.

$\endgroup$ 6 Sorted by: Reset to default

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