Question

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

Answer 1

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,
$\underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{x^2-6x+9}}{x-3}}$
The numerator consists of a quadratic equation, now we will simplify it as follows,
$\underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{x^2-3x-3x+9}}{x-3}}$
$\underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{x(x-3)-3(x-3)}}{(x-3)}}$
$$\Rightarrow \underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{(x-3)(x-3)}}{(x-3)}}$$
$\Rightarrow \underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{{(x-3)}^2}}{(x-3)}}$
We know, $\sqrt{4}=\pm 2$ , so the above equation can be written as,
$\Rightarrow \underset{x\to 3}{lim}f(x)=\underset{x\to 3}{lim}{\displaystyle \frac{|x-3|}{(x-3)}}.......(i)$
Now the modulus can be split as following,
$f(x)=\{\begin{array}{c}{\displaystyle \frac{x-3}{x-3}},x>0\\ {\displaystyle \frac{-(x-3)}{x-3}},x<0\end{array}$
Now we will find the left hand limit, we get
$\underset{x\to 3^{-}}{lim}f(x)=\underset{x\to 3^{-}}{lim}{\displaystyle \frac{-(x-3)}{x-3}}$
Cancelling the like terms, we get
$\underset{x\to 3^{-}}{lim}f(x)=-1$
So, the left hand limit of the given function is ${}^{\prime }-1^{\prime }$.
Now we will find the right hand limit, we get
$\underset{x\to 3^{+}}{lim}f(x)=\underset{x\to 3^{+}}{lim}{\displaystyle \frac{(x-3)}{x-3}}$
Cancelling the like terms, we get
$\underset{x\to 3^{+}}{lim}f(x)=1$
So, the right hand limit of the given function is ${}^{\prime }{1}^{\prime }$.
So the left hand limit and the right hand limit are not equal hence the $\underset{x\to 3}{lim}f(x)$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 $$\underset{x\to 0^{-}}{lim}g(x)=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{g(0+h)-g(0)}{h}}$$, this formula, but it will be complicated process.