Show that sin x is Continuous: Proof

In this post, we will show that the trigonometric function sinx is continuous for all values of x. Here, we will use the limit method as well as the epsilon-delta definition.

sinx is continuous

Note that

$\lim\limits_{x \to c} \sin x=\sin c$ and $\sin x$ is defined for all real numbers. Thus, we can say that the function $\sin x$ is continuous everywhere.

Now, we will show that $\sin x$ is continuous at every real number by the epsilon-delta method.

Table of Contents

Prove that sinx is Continuous

To prove the sine function is continuous by the epsilon-delta definition, we will use the following two formulas:

  1. $\sin x -\sin y=2\sin \dfrac{x-y}{2} \cos \dfrac{x+y}{2}$
  2. $|\sin x| \leq |x|$

Let $f(x)=\sin x$ and $x=c$ be an arbitrary point. Let ε>0 be any given positive number. We need to find a positive δ such that

|f(x)-f(c)| < ε whenever 0<|x-c|<δ

Let us choose $\delta=\epsilon$. Now, for $x \in {0<|x-c| <\delta}$, we have that

$|f(x)-f(c)| = |\sin x -\sin c|$

$=2|\sin \dfrac{x-c}{2} \cos \dfrac{x+c}{2}|$ by the above Formula 1.

$\leq 2 |\sin \dfrac{x-c}{2}|$ as we know that $|\cos x| \leq 1$ for all values of x.

$\leq 2|\dfrac{x-c}{2}|$by the above Formula 2.

$=|x-c|$

$<\epsilon$.

Thus, we have shown that for any given $\epsilon>0$, there exists a $\delta>0$ such that whenever $0<|x-c|<\delta$, we have

$|f(x)-f(x)|<\epsilon$.

Hence, by the epsilon-delta definition of Continuity, we conclude that the function $f(x)=\sin x$ is continuous at $x=c$. As the point $c$ was arbitrary, $\sin x$ is everywhere continuous.

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FAQs

Q1: Is sinx continuous?

Answer: Yes, sinx is continuous for all real values x.

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