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

$\begin{array}{1 1} \text{Yes it is continuous at x }= \pi \\ \text{No, it is not continuous at} \;x = \pi \end{array} $

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL =$\lim\limits_{\large x\to \pi}(x^2-\sin x+5)$
Put $x=\pi-h$
$\Rightarrow \lim\limits_{\large h\to 0}[(\pi-h)^2-\sin(\pi-h)+5]$
$\Rightarrow \lim\limits_{\large h\to 0}\pi^2-2\pi h+h^2-\sin h+5]$
$\Rightarrow \pi^2+5$
Step 2:
RHL=$\lim\limits_{\large x\to \pi^+}(x^2-\sin x+5)$
Put $x=\pi+h$
$\Rightarrow \lim\limits_{\large h\to 0}[(\pi+h)^2-\sin(\pi+h)+5]$
$\Rightarrow \lim\limits_{\large h\to 0}\pi^2+2\pi h+h^2+\sin h+5]$
$\Rightarrow \pi^2+5$
Hence $f$ is continuous at $x=\pi$
Step 3:
Alternatively $g(x)=x^2+5$ is a polynomial and $h(x)=\sin x$
Therefore $g$ is continuous for all $x\in R$
$f=g-h$ is also continuous for $x\in R$
answered May 28, 2013 by sreemathi.v

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