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Is the function f defined by $f(x) = \left\{\begin{array}{l l} x, & \quad \text{if $x$ \( \leq \) 1}\\ 5, & \quad \text{if $x$ > 1}\\ \end{array} \right.$ continuous at $(x = 0)$? At $(x = 1)$? At $(x = 2)$?

$\begin{array}{1 1} \text{Yes, continuous at x = 0 ,1 and 2}\\ \text{Yes, continuous at x = 0 and 1 and discontinuous at x = 2} \\ \text{Yes, continuous at x = 1 and 2 and discontinuous at x = 0} \\ \text{Yes, continuous at x = 0 and 2 and discontinuous at 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 $x=0$
$\lim\limits_{\large x\to 0}f(x)=\lim\limits_{\large x\to 0^-}x=0$
$\lim\limits_{\large x\to 0}f(x)=\lim\limits_{\large x\to 0^+}x=0$
$f(0)=0$
$f$ is continuous at $x=0$
Step 2:
At $x=1$
$\lim\limits_{\large x\to 1^-}f(x)=\lim\limits_{\large x\to 1^-}x=1$
$\lim\limits_{\large x\to 1^+}f(x)=\lim\limits_{\large x\to 1^+}x=5$
$\lim\limits_{\large x\to 1^-}f(x)\neq\lim\limits_{\large x\to 1^+}f(x)$
$f$ is discontinuous at $x=1$
Step 3:
At $x=2$
$\lim\limits_{\large x\to 2}f(x)=5$
$f(2)=5$
$f$ is continuous at $x=2$
answered May 27, 2013 by sreemathi.v
 

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