91Ó°ÊÓ

For a function to be continuous at a point \(x_0\), the limit must exist, and the limit as \(x\) approaches \(x_0\) must be equal to the function's value at \(x_0\). Mathematically, this can be written as: \[ \lim_{x\to x_0} f(x) = f(x_0) \] Our goal is to show that this condition holds for any \(x_0 \in \mathbb{R}\).

We can rewrite the limit in terms of the difference between the function values at any given point \(x\) and the point of interest \(x_0\). In order to do this, we will use the difference identity for cosine: \[\cos{A} - \cos{B} = -2\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A - B}{2}\right)\] Let \(A = x\) and \(B = x_0\), then we can write the difference of the function values as: \[\cos x - \cos x_0 = -2\sin \left(\frac{x+x_0}{2}\right) \sin \left(\frac{x-x_0}{2}\right)\]

Now, we rewrite the limit in terms of the difference identity: \[\lim_{x\to x_0} \left(\cos x - \cos x_0\right) = \lim_{x\to x_0} \left[ -2\sin \left(\frac{x+x_0}{2}\right) \sin \left(\frac{x-x_0}{2}\right) \right]\] Notice that the only part of the expression which depends on \(x\) and approaches 0 as \(x \to x_0\) is the term \(\sin \left(\frac{x - x_0}{2}\right)\). We will use the fact that \(\lim_{x\to 0} \frac{\sin x}{x} = 1\), a well-known limit. Now let us make the substitution \(u = x - x_0\), which implies that \(x \to x_0\) when \(u \to 0\). Thus, the limit becomes \[\lim_{u \to 0} \frac{-2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \sin \left(\frac{u}{2}\right)}{u}\] This limit can be separated into two: \[\lim_{u \to 0} \left[ -2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \cdot \lim_{u \to 0} \frac{\sin \left(\frac{u}{2}\right)}{u} \right]\]

Next, apply the known limit \(\lim_{x\to 0} \frac{\sin x}{x} = 1\) to the second limit and simplify: \[\lim_{u \to 0} \left[ -2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \cdot 1 \right] = -2\, \sin \left(\frac{x_0 + x_0}{2}\right) \cdot 0\] The entire expression evaluates to 0: \[\lim_{x\to x_0} (\cos x - \cos x_0) = 0\] This means that the limit of the difference between any two points of the function as they approach each other is 0, which implies that the function is continuous at any point \(x_0\). Therefore, the function \(f(x) = \cos x\) is continuous everywhere.