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Examine the continuity of f(x) at the indicated point $f(x)=\left \{\begin{array}{1 1}3x+5 & if\;x\geq 2\\x^2 & if\;x<2\end{array}\right.$ at $x=2.$

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Toolbox:
  • A function is said to be discontinuous at $x=a$,if both the right hand limit and left hand limit does not exist.
  • It is also discontinuous if RHL $\neq$ LHL.
Step 1:
$f(x)=3x+5, \quad x \geq 2$
$f(2)=3(2)+5$
$\qquad=6+5$
$\qquad=11$
Step 2:
RHL: In this case $x \geq 2$
$f(2^+)=\lim\limits _{\large x \to 2+} f(x)=\lim \limits_{\large x \to 2^+} (3x+5)$
On applying limits we get
$\qquad\quad\qquad\qquad\;\; =3(2)+5$
$\qquad\quad \qquad\qquad\;\;=11$
Step 3:
LHL: In this case $x < 2$
$f(2-)=\lim\limits _{\large x \to 2^-} f(x)=\lim\limits _{\large x \to 2^-} x^2$
On applying limits we get
$\qquad\quad \qquad\qquad\;\;=(2)^2$
$\qquad\quad \qquad\qquad\;\;=4$
Step 4:
Hence $\lim\limits_{\large x\to 2^-}f(x) \neq \lim\limits_{\large x\to 2^+}f(x)$
Therefore $f(x) $ is not continuous at $x=2$
answered Jun 25, 2013 by sreemathi.v
 

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