Discuss the continuity of the following functions :

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

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Solution

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

= $\sqrt{2} s i n (\frac{1}{\sqrt{2}} s i n x - \frac{1}{\sqrt{2}} c o s x) = \sqrt{2} s i n (s i n x c o s \frac{π}{4} - c o s x s i n \frac{π}{4})$

$= \sqrt{2} s i n (x - \frac{π}{4})$

At x=a a, where $a ϵ R$

LHL = $l i m x \to a^{-} f (x) = l i m x \to 0^{-} \sqrt{2} s i n (x = \frac{π}{4}) l i m h \to 0 \sqrt{2} s i n (a - h + \frac{π}{4})$

= $l i m x \to 0^{-} \sqrt{2} [s i n (a + \frac{π}{4}) c o s h - c o s (a + \frac{π}{4}) s i n h]$

= $\sqrt{2} [s i n (a - \frac{π}{4})] c o s 0 - \sqrt{2} c o s (a - \frac{π}{4}) s i n 0 = \sqrt{2} s i n (a - \frac{π}{4})$

RHL = $l i m x \to a^{+} \sqrt{2} s i n (x = \frac{π}{4}) l i m h \to 0 \sqrt{2} s i n (a - h + \frac{π}{4})$

$l i m x \to 0^{-} = \sqrt{2} [s i n (a - \frac{π}{4}) c o s h + c o s (a - \frac{π}{4}) s i n h]$

[Use formula sin (A+B)]

= $\sqrt{2} [s i n (a - \frac{π}{4})] c o s 0 + \sqrt{2} c o s (a - \frac{π}{4}) s i n 0 = \sqrt{2} s i n (a - \frac{π}{4})$

Also, f(a) = $\sqrt{2} s i n (a - \frac{π}{4})$

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

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