Question

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

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.

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.