Question

How do you find the limit of $\dfrac{x-3}{\left| x-3 \right|}$ as x approaches 3?

Answer 1

Hint: To find the limit of $\dfrac{x-3}{\left| x-3 \right|}$ as x approaches 3, we will use the function of modulus. We know that for an absolute value of a variable, say x, we can see that $\left| x \right|=x$ , if $x\ge 0$ and $\left| x \right|=-x$ , if $x<0$ . We will then find the limit for each case, that is, $\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)$ and $\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)$ . If both the limits are equal, then the limit of the given function exists and will be equal to the value obtained. If both the limits are not equal, then the limit of the given function does not exist.

Complete step by step solution:
We have to find the limit of $\dfrac{x-3}{\left| x-3 \right|}$ as x approaches 3. We know that for an absolute value of a variable, say x, we can see that $\left| x \right|=x$ , if $x\ge 0$ …(i)
and $\left| x \right|=-x$ , if $x<0$ …(ii)
Let us denote the given expression as $f\left( x \right)=\dfrac{x-3}{\left| x-3 \right|},x\ne 3$ .
As $x \to {{3}^{+}}$ , this means that $x>3$ . Let us take 3 from RHS to LHS.
$\begin{align}
  & x>3 \\
 & \Rightarrow x-3>0 \\
\end{align}$
From (i), we can write $\left| x-3 \right|=x-3$ .
Let us find the value of the function for this case.
$f\left( x \right)=\dfrac{x-3}{\left| x-3 \right|}=\dfrac{x-3}{x-3}=1,x\ne 3$
Now, let us take the limit of the above function.
$\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}1$
We know that the limit of a constant is constant itself.
$\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=1...\left( a \right)$
Now let us consider the case when $x \to {{3}^{-}}$ . This means that $x<3$ . Let us take 3 from RHS to LHS.
$\begin{align}
  & x<3 \\
 & \Rightarrow x-3<0 \\
\end{align}$
From (ii), we can write $\left| x-3 \right|=-\left( x-3 \right)$ .
Let us find the value of the function for this case.
$f\left( x \right)=\dfrac{x-3}{\left| x-3 \right|}=\dfrac{x-3}{-\left( x-3 \right)}=-1,x\ne 3$
Now, let us take the limit of the above function.
$\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\left( -1 \right)$
We know that the limit of a constant is constant itself.
$\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=-1...\left( b \right)$
From equations (a) and (b), we can see that
$\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)\ne \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)$
Hence, $\displaystyle \lim_{x \to 3}f\left( x \right)$ limit does not exist.

Note: Students must know the function of modulus clearly before solving these problems. They must also know how to apply limits and the rules in limits. If $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\ne \displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)$ , then the limit does not exists. If $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)$ , then the limit exists and will be equal to the value obtained.