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Question

Discuss the continuity of the function: f(x) = sin x - cos x

Vidyadip · Accepted Answer

We known that if g and r are two continuous functions, then
g + r, g – r and g.r are also continuous.
First we have to prove that g(x) = sinx and r(x) = cosx are continuous functions.
Now, let g(x) = sinx
We know that g(x) = sinx is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
g(h) = sin h
$\mathop {\lim }\limits_{x \to h} g(x) = \mathop {\lim }\limits_{x \to h} \sin x$
= $\mathop {\lim }\limits_{k \to 0} \sin (h + k)$
= $\mathop {\lim }\limits_{k \to 0} [\sinh \cos k + \cosh \sin k]$
= sinh.cos0 + cosh.sin0
= sinh + 0
= sin h
Thus $\mathop {\lim }\limits_{x \to h} g(x) = g(h)$
Therefore, g is a continuous function …(1)
Now, let f(x) = cos x
We know that f(x) = cos x is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
Now f(h) = cosh
$\mathop {\lim }\limits_{x \to h} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{h}}} \cos {\text{x}}$
= $\mathop {\lim }\limits_{{\text{k}} \to 0} \cos ({\text{h}} + {\text{k}})$
= $\mathop {\lim }\limits_{x \to 0} [\cosh \cos k - \sinh \sin k]$
= coshcos0 - sinhsin0
= cosh - 0
= cosh
Thus $\mathop {\lim }\limits_{x \to h} f(x) = f(h)$
Therefore, f is a continuous function ….(2)
So, from (1) and (2), we get,
r(x) = g(x) - f(x) = sinx - cosx is a continuous function.

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