Question

Prove that sine function is continuous at every real number.

Answer 1

Let $f\left(x\right)=sinx$

Let c be any real number.

We know that A function is continuous at $x=c$

If L.H.L = R.H.L= f(c)

i.e. $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c^{+}}{lim}f\left(x\right)=f\left(c\right)$

Taking L.H.L

$\underset{x\to c^{-}}{lim}f\left(x\right)$

$\underset{x\to c^{-}}{lim}\left(sinx\right)$

since sin x is defined for every real number.

Putting $x=c-h$

$x\to c^{-}$

$c-h\to x$

$-h\to 0$

$h\to 0$

$=\underset{h\to 0}{lim}\sin (c-h)$

$=\underset{h\to 0}{lim}(\sin ccosh-\sin c\sin h)$

putting h = 0

=$sinccos0-cosc.sin0$

=$sinc\left(1\right)-cosc.0$

= $sinc$

Taking R.H.L

$\underset{x\to c^{+}}{lim}f\left(x\right)$

$\underset{sinx\to c^{+}}{lim}sin\left(x\right)$

putting $x=c+h$

$\underset{h\to 0}{lim}\sin (c+h)$

$\underset{h\to 0}{lim}\sin (sinc\cos h+coscsinh)$

putting $h=0$

$=sinccos0+cosc.sin0$

$=sin\left(1\right)+cosc.0$

$=sinc$

$f\left(x\right)=sinx$

$f\left(c\right)=sinc$

Hence $L.H.L=R.H.L=f\left(c\right)$

$\underset{x\to c^{-}}{lim}f\left(x\right)=\underset{x\to c^{+}}{lim}f\left(x\right)=f\left(c\right)$

$f\left(x\right)$ is continuous

so, is continous.