A) 0
B) 3
C) \[\infty \]
D) 6
E) does not exist
Correct Answer: E
Solution :
\[\underset{x\to 3}{\mathop{\lim }}\,\frac{({{x}^{2}}-9)}{|x-3|}\] Now, \[|x-3|=\left\{ \begin{matrix} x-3, & if\,x>3 \\ -(x-3), & if\,x<3 \\ \end{matrix} \right.\] \[\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{|x-3|}=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\frac{(x+3)(x-3)}{(x-3)}=6\] and\[\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{|x-3|}=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\frac{(x+3)(x-3)}{-(x-3)}=-6\] \[\therefore \] Limit does not exist.
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