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Discuss the continuity of the function \(f\), where \(f\) is defined by $f(x) = \left\{ \begin{array} {1 1} -2 ,& \quad\text{ if $ x $ \(\leq -1\)}\\ 2x,& \quad\text{ if $ -1 $ < x \(\leq1\)}\\ 2 ,& \quad\text{ if $ x $ > 1}\\ \end{array} \right. $

$\begin{array}{1 1} \text{Yes, f is continuous at x=1 and x= -1} \\ \text{ No f is not continuous at x=1 and x= -1} \end{array} $

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Toolbox:
  • 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:
At $x=-1$
LHL=$\lim\limits_{\large x\to -1}f(x)=-2$
RHL=$\lim\limits_{\large x\to -1}f(x)=-2$
Therefore LHS=RHS.
$\Rightarrow f$ is continuous at $x=-1$
Step 2:
At $x=1$
LHL=$\lim\limits_{\large x\to 1^-}f(x)=2x$
$\qquad\qquad\qquad=2$
RHL=$\lim\limits_{\large x\to 1^+}f(x)=2x$
$\qquad\qquad\qquad=2$
Therefore LHS=RHS.
$\Rightarrow f$ is continuous at $x=1$
answered May 28, 2013 by sreemathi.v
 

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