# Discuss the continuity of the following functions $(b)\;f(x)=\sin x-\cos x$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• $\sin(a-b)=\sin a\cos b-\cos a\sin b$
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$f(x)=\sin x-\cos x$
Multiply and divide by $\sqrt 2$
$\qquad=\sqrt 2[\large\frac{1}{\sqrt 2}$$\sin x-\large\frac{1}{\sqrt 2}$$\cos x]$
$\qquad=\sqrt 2[\sin x\cos\large\frac{\pi}{4}$$-\cos x\sin\large\frac{\pi}{4}] The above equation is of the form \sin(a-b)=\sin a\cos b-\cos a\sin b \Rightarrow \sqrt 2\sin(x-\large\frac{\pi}{4}) Step 2: At x=a where a\in R LHL=\lim\limits_{\large x\to a^-}\sqrt 2\sin(x-\large\frac{\pi}{4}) \qquad=\lim\limits_{h\to 0}\sqrt 2\sin(a-h-\large\frac{\pi}{4}) \qquad=\lim\limits_{h\to 0}\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos h-\cos(a-\large\frac{\pi}{4})$$\sin h \sin(A-B)=\sin A\cos B-\cos A\sin B \qquad=\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos 0-\sqrt 2\cos(a-\large\frac{\pi}{4})$$\sin 0 \qquad=\sqrt 2\sin(a-\large\frac{\pi}{4}) Step 3: RHL=\lim\limits_{\large x\to a^+}\sqrt 2\sin(x-\large\frac{\pi}{4}) \qquad=\lim\limits_{h\to 0}\sqrt 2\sin(a+h-\large\frac{\pi}{4}) \qquad=\lim\limits_{h\to 0}\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos h+\cos(a-\large\frac{\pi}{4})$$\sin h \qquad=\sqrt 2[\sin(a-\large\frac{\pi}{4})$$\cos 0-\cos(a-\large\frac{\pi}{4})$$\sin 0]$
$\qquad=\sqrt 2\sin(a-\large\frac{\pi}{4})$
Step 4:
Also $f(a)=\sqrt 2\sin(a-\large\frac{\pi}{4})$
$\therefore$ LHL =RHL=f(a).
Hence $f(x)$ is continuous at all points.