Question
Is the function defined by $f\left( x \right) = {x^2} - \sin x + 5$ continuous at $x = \pi $?
L.H.L. $= \mathop {\lim }\limits_{x \to {\pi ^ - }} \left( {{x^2} - \sin x + 5} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {\pi - h} \right)}^2} - \sin \left( {\pi - h} \right) + 5} \right] = {\pi ^2} + 5$
R.H.L.$= \mathop {\lim }\limits_{x \to {\pi ^ + }} \left( {{x^2} - \sin x + 5} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {\pi + h} \right)}^2} - \sin \left( {\pi + h} \right) + 5} \right] = {\pi ^2} + 5$
And $f\left( \pi \right) = {\pi ^2} - \sin \pi + 5 = {\pi ^2} + 5$
Since L.H.L. = R.H.L. = $f\left( \pi \right)$
Therefore, f is continuous at $x = \pi $
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