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20. Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = \pi?

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The given function is
f (x) = x^2 - sin x + 5
Clearly, given function is defined at x =\pi
f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = f(\pi)
Hence, the function defined by f (x) = x^2 - sin x + 5 continuous at x = \pi

Posted by

Gautam harsolia

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