Real Sine Function is Continuous

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Theorem

Let $x \in \R$ be a real number.

Let $\sin x$ be the sine of $x$.


Then:

$\sin x$ is continuous on $\R$.


Proof

Recall the definition of the sine function:

$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

Thus $\sin x$ is expressible in the form of a power series.

From Sine Function is Absolutely Convergent, we have that the interval of convergence of $\sin x$ is the whole of $\R$.

From Power Series is Differentiable on Interval of Convergence, it follows that $\sin x$ is continuous on the whole of $\R$.

$\blacksquare$