Find the value

Question:

Let $f(x)=\left\{\begin{array}{l}\frac{|x-3|}{(x-3)^{\prime}} x \neq 3 \\ 0, \quad x=3\end{array}\right.$

Show that $\lim _{x \rightarrow 3} f(x)$ does not exist.

 

Solution:

Left Hand Limit(L.H.L.):

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} \frac{|x-3|}{x-3}$

$=\lim _{x \rightarrow 3^{-}} \frac{-(x-3)}{x-3}$

$=\lim _{x \rightarrow 3^{-}}-1$

$=-1$

Right Hand Limit(R.H.L.):

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} \frac{|x-3|}{x-3}$

$=\lim _{x \rightarrow 3^{+}} \frac{(x-3)}{x-3}$

$=\lim _{x \rightarrow 3^{+}} 1$

$=1$

$\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$

Thus, $\lim _{x \rightarrow 3} f(x)$ does not exist.

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now