91Ó°ÊÓ

Recall the limit definition: If \[ \lim _{x \rightarrow a} f(x) = L\], then for every ε > 0, there exists a δ > 0, such that if 0 < |x - a| < δ, then |f(x) - L| < ε. In our case, we have to show that for every ε > 0, there exists a δ > 0, such that if 0 < |x - a| < δ, then |cos(x) - cos(a)| < ε.

To proof the limit, let's use the auxiliary function g(x) = cos(x), then \[|g(x) - g(a)| = |cos(x) - cos(a)|\]

We can rewrite |cos(x) - cos(a)| using a trigonometric identity that involves the sine function, which is: \[|cos(x) - cos(a)| = |-2 sin(\frac{x - a}{2}) sin(\frac{x + a}{2})|\] Keep in mind that the sine function is always between -1 and 1, so we can apply the inequality \(-1 \leq sin(t) \leq 1\)

We apply inequality for sine function, and thus have \(-2 |sin(\frac{x - a}{2}) sin(\frac{x + a}{2})| \leq -2 sin(\frac{x - a}{2}) sin(\frac{x + a}{2}) \leq 2 |sin(\frac{x - a}{2}) sin(\frac{x + a}{2})|\) Now, for every ε > 0, there exists δ > 0, such that if 0 < |x - a| < δ, then \(-2 ε \leq -2 sin(\frac{x - a}{2}) sin(\frac{x + a}{2}) \leq 2 ε\) Since sin(x) is continuous and differentiable, we know that \(\lim _{x \rightarrow 0} sin(x) = 0\). This means that \[\lim _{x \rightarrow a} sin(\frac{x - a}{2}) sin(\frac{x + a}{2}) = 0\] As a result, we can choose δ > 0 such that \[ -2 sin(\frac{x - a}{2}) sin(\frac{x + a}{2}) \leq -2 ε\] and \[2 sin(\frac{x - a}{2}) sin(\frac{x + a}{2}) \leq 2 ε\] Since we have proven these inequalities for every ε, it proves that \[\lim_{x \rightarrow a} \cos x = \cos a\]