Question

Discuss the continuity of the following functions :

(a) f(x) = sin x + cos x

(b) f(x) = sin x + cos x

(c) f(x) = sin x cos x

Answer 1

(a) f(x) =sin x + cos x $=\sqrt{2}(\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx)$

$=\sqrt{2}(cos\frac{\pi }{4}sinx+cosxsin\frac{\pi }{4})=\sqrt{2}sin(x+\frac{\pi }{4})$

At x = a, where $aϵR$

$LHL=\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}\sqrt{2}sin(x+\frac{\pi }{4})=\underset{h\to 0}{lim}\sqrt{2}sin(a-h+\frac{\pi }{4})$

$=\sqrt{2}\underset{h\to 0}{lim}[sin(a+\frac{\pi }{4})cosh-cos(a+\frac{\pi }{4})sinh]$

[$\therefore$ sin(A-B)=sin A cos B - cos A sin B]

$=\sqrt{2}sin(a+\frac{\pi }{4})cos0-\sqrt{2}cos(a+\frac{\pi }{4})sin0=\sqrt{2}sin(a+\frac{\pi }{4})$

$RHL=\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\sqrt{2}sin(x+\frac{\pi }{4})=\underset{h\to 0}{lim}\sqrt{2}sin(a+h+\frac{\pi }{4})$

$=\sqrt{2}\underset{h\to 0}{lim}[sin(a+\frac{\pi }{4})cosh+cos(a+\frac{\pi }{4})sinh]$

[$\because$ sin(A+B)=sin A cos B - cos A sin B]

$=\sqrt{2}sin(a+\frac{\pi }{4})cos0-sin0cos(a+\frac{\pi }{4})$

=$\sqrt{2}sin(a+\frac{\pi }{4})$

Also, f(a)=$\sqrt{2}sin(a+\frac{\pi }{4})$

$\therefore$ LHL = RHL = f(a)

Hence, f(x) is continuous at all points.

Here, f(x) = sin x-cos x

=$\sqrt{2}sin(\frac{1}{\sqrt{2}}sinx-\frac{1}{\sqrt{2}}cosx)=\sqrt{2}sin(sinxcos\frac{\pi }{4}-cosxsin\frac{\pi }{4})$

$=\sqrt{2}sin(x-\frac{\pi }{4})$

At x=a , where $aϵR$

LHL = $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}\sqrt{2}sin(x-\frac{\pi }{4})=\underset{h\to 0}{lim}\sqrt{2}sin(a-h-\frac{\pi }{4})$

=$\underset{h\to 0}{lim}\sqrt{2}[sin(a-\frac{\pi }{4})cosh-cos(a-\frac{\pi }{4})sinh]$

=$\sqrt{2}[sin(a-\frac{\pi }{4})cos0-cos(a-\frac{\pi }{4})sin0]=\sqrt{2}sin(a-\frac{\pi }{4})$

RHL = $\underset{x\to a^{+}}{lim}\sqrt{2}sin(x-\frac{\pi }{4})=\underset{h\to 0}{lim}\sqrt{2}sin(a+h-\frac{\pi }{4})$

$\underset{h\to 0}{lim}=\sqrt{2}[sin(a-\frac{\pi }{4})cosh+cos(a-\frac{\pi }{4})sinh]$

[Use formula sin (A+B)]

=$\sqrt{2}\left[sin(a-\frac{\pi }{4})\right]cos0+\sqrt{2}cos(a-\frac{\pi }{4})sin0=\sqrt{2}sin(a-\frac{\pi }{4})$

Also, f(a) =$\sqrt{2}sin(a-\frac{\pi }{4})$

$\therefore$ LHL = RHL = f(a). Hence, f(x) is continuous at all points.

Here, f(x) =sin x cos x

f(x)=$\frac{1}{2}\times 2sinxcosx=\frac{1}{2}sin2x$

At x=a, where $aϵR$

LHL = $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}\frac{1}{2}sin2x=\underset{h\to 0}{lim}\frac{1}{2}sin2(a-h)$

=$\underset{h\to 0}{lim}\frac{1}{2}[sin2acos2h-cos2asin2h]$

=$\frac{1}{2}[sin2acos0-cos2asin0]=\frac{1}{2}sin2a\left[\text{Use formula sin(A+B)}\right]$

RHL = $\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}\frac{1}{2}sin2x=\underset{h\to 0}{lim}\frac{1}{2}sin2(a+h)$

=$\underset{h\to 0}{lim}\frac{1}{2}[sin2acos2h-cos2asin2h]\left[\text{Use formula sin(A-B)}\right]$

=$\frac{1}{2}sin2a$

Also, $f\left(a\right)=\frac{1}{2}sin2a$

Therefore LHL=RHL=f(a).
Hence, f(x) is continuous at all points.