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Examine the continuity of the function $f (x) = 2x^2 - 1$ at $x = 3$

$\begin{array}{1 1}\text{f(x) is continuous at x=3} \\ \text{f(x) is NOT continuous at x=3} \end{array} $

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  • If $f$ is a real function on a subset of the real numbers and $c$ a point in the domain of $f$, then $f$ is continous at $c$ if $\lim\limits_{x\to c} f(x) = f(c)$.
Given $f(x) = 2x^2-1$.
At $x=3, \; \lim\limits_{x\to 3} f(x) = \lim\limits_{x\to 3} 2x^2-1 = 2 \times 3^2 -1 = 2 \times 9 - 1= 18-1 = 17$
$f(3) = 2 \times 3^2 -1 = 2 \times 9 - 1= 18-1 = 17$
Since $\lim\limits_{x\to 3} f(x) = f(3), f(x)$ is continous at $x=3$.
answered Apr 4, 2013 by balaji.thirumalai
 
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