Question

Discuss the continuity of sine function.

Answer 1

To find the continuity of given function, we do it by the method of limits. First we will find the left hand limit at any real point of the function, then we find the right hand limit at that real point of the function and then we find the value of that function at that real point. If the value of all three of them is equal, we say that the function is continuous at that real point.
Formula used: A function $f(x)$ is said to continuous if it follows the condition,
$LHL = RHL = f(c)$
i.e. $\mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = f(c)$

Complete step-by-step solution:
We are given a sine function. Let $f(x) = \sin x$
We will check the continuity of $f(x)$ at any real number let $c$.
We know that a function is continuous at $x = c$ if
LHL=RHL= $f(c)$
So first we find LHL at $c$.

$$\mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(c - x) \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin (c - x) \\\$$

We know that $\sin (x - y) = \sin x\cos y - \cos x\sin y$, using this we get,

$$\Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} (\sin c\cosh - \cos c\sinh ) \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \sin c\cos 0 - \cos c\sin 0 \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \sin c \times 1 - 0 \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \sin c \\\$$

Now we find the RHL,

$$\mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(c + x) \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin (c + x) \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} (\sin c\cosh + \cos c\sinh ) \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \sin c\cos 0 + \cos c\sin 0 \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \sin c \times 1 + 0 \\\ \Rightarrow \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \sin c \\\$$

Now we find the value of sine function for $c$ as,
$f(c) = \sin c$
Since $LHL = RHL = f(c)$, we say that sine function is continuous.

Note: When we have to prove any function continuous, we have to prove all the above given conditions equal. Even if any of the one condition is different we say that the function is not continuous. For the extreme left point, LHL can be different and we call that function to be right continuous at that point and for the extreme right point, RHL can be different and we call that function to be left continuous at that point.