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Find all points of discontinuity of \(f\), where \(f\) is defined by $ f(x) = \left\{ \begin{array} {1 1} x^{10} -1 ,& \quad\text{ if $ x $ \(\leq 1\)}\\ x^2,& \quad \text{if $x$ > 1}\\ \end{array} \right. $

$\begin{array}{1 1} \text{Point of discontinuity is x=1} \\ \text{Point of discontinuity is x=-1} \\ \text{Points of discontinuity are all x<1} \\ \text{Points of discontinuity are all x > 1}\end{array} $

1 Answer

  • 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 =$\lim\limits_{\large x\to 1^-}f(x)=(x^{10}-1)$
RHL =$\lim\limits_{\large x\to 1^+}f(x)=(x^2)$
$\Rightarrow $LHL $\neq$ RHL $\neq$ f(1)
$f$ is not continuous at $x=1$
Step 2:
At $x=c < 1$
$\lim\limits_{\large x\to c}(x^{10}-1)=(c^{10}-1)$
At $x=c > 1$
$\lim\limits_{\large x\to c}(x^2)=(c^2)$
Step 3:
$f$ is continuous at all points $x\in R-\{1\}$
Therefore point of discontinuity is $x=1$
answered May 28, 2013 by sreemathi.v
edited Jul 1, 2014 by rvidyagovindarajan_1

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