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Evaluate the left hand and right hand limits of the function\[f(x)=\left\{ \begin{matrix} \dfrac{\sqrt{({{x}^{2}}-6x+9)}}{(x-3)},x\ne 3 \\ 0,x=3 \\
\end{matrix}\text{ at }x=3. \right.\]

Answer
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Hint: For finding out whether the limit exists, then we should find the left hand limit and right hand limit. If they are equal then the limit exists.

Complete step-by-step answer:
First we will simplify the given function, that is,
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{x}^{2}}-6x+9}}{x-3}$
The numerator consists of a quadratic equation, now we will simplify it as follows,
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{x}^{2}}-3x-3x+9}}{x-3}$
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{x(x-3)-3(x-3)}}{(x-3)}$
\[\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{(x-3)(x-3)}}{(x-3)}\]
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{(x-3)}^{2}}}}{(x-3)}$
We know, $\sqrt{4}=\pm 2$ , so the above equation can be written as,
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\left| x-3 \right|}{(x-3)}.......(i)$
Now the modulus can be split as following,
$f(x)=\left\{ \begin{matrix}
   \dfrac{x-3}{x-3},x>0 \\
   \dfrac{-(x-3)}{x-3},x<0 \\
\end{matrix} \right.$
Now we will find the left hand limit, we get
$\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\dfrac{-(x-3)}{x-3}$
Cancelling the like terms, we get
$\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=-1$
So, the left hand limit of the given function is $'-1'$.
Now we will find the right hand limit, we get
$\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\dfrac{(x-3)}{x-3}$
Cancelling the like terms, we get
$\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=1$
So, the right hand limit of the given function is $'1'$.
So the left hand limit and the right hand limit are not equal hence the $\underset{x\to 3}{\mathop{\lim }}\,f(x)\text{ }$does not exist.

Note: Generally these questions are asked in competitive examinations for confusing students. Instead of solving and then finding the left hand limit and right hand limit. We can directly apply the \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(x)=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{g(0+h)-g(0)}{h}\], this formula, but it will be complicated process.