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Show that the function f(x) = |sin x + cos x| is continuous at x = $\pi$ .

Answers (1)

Given,

$f(x)=|\sin x+\cos x| \underline{\ldots}(1)$

We need to prove that f(x) is continuous at x = π

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = π if -

$\lim _{h \rightarrow 0} f(\pi-h)=\lim _{h \rightarrow 0} f(\pi+h)=f(\pi)$

Now,

LHL = $\lim _{h \rightarrow 0} f(\pi-h)$

⇒ LHL = $\lim _{h \rightarrow 0}\{|\sin (\pi-h)+\cos (\pi-h)|\}$ {using eqn 1}

$\because \sin (\pi-x)=\sin x \& \cos (\pi-x)=-\cos x$

$\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}|\sinh -\cosh |$

$\Rightarrow \mathrm{LHL}=|\sin 0-\cos 0|=|0-1|$

$\therefore \mathrm{LHL}=1 \underline{\ldots(2)}$

Similarly, we proceed for RHL-

$\operatorname{RHL}=\lim _{h \rightarrow 0} f(\pi+h)$

$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\{|\sin (\pi+\mathrm{h})+\cos (\pi+\mathrm{h})|\}$ {using eqn 1}

$\because \sin (\pi+x)=-\sin x \& \cos (\pi+x)=-\cos x$

$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|-\sin \mathrm{h}-\cosh |$

$\Rightarrow \mathrm{RHL}=|-\sin 0-\cos 0|=|0-1|$

$\therefore \mathrm{RHL}=1 \ldots(3)$

$\text { Also, } f(\pi)=|\sin \pi+\cos \pi|=|0-1|=1 \underline{\ldots(4)}$

Now from equation 2, 3 and 4 we can conclude that

$\lim _{h \rightarrow 0} \mathrm{f}(\pi-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(\pi+\mathrm{h})=\mathrm{f}(\pi)=1$

∴ f(x) is continuous at x = π is proved

Posted by

infoexpert21

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