Question

If f is a function defined by $f\left(x\right)=\{\begin{array}{ccc}\frac{x-1}{\sqrt{x}-1}& if& x>1\\ 5-3x& if& -2\le x\le 1,\\ \frac{6}{x-10}& if& x<-2\end{array}$ then discuss the continuity of f.

Answer 1

For the continuity of f at $x=1$

$RHL=\underset{x\to 1}{lim}\frac{x-1}{\sqrt{x}-1}$

Using L's Hospital rule, we get

$\underset{x\to 1}{lim}2\sqrt{x}=2$

$LHL=\underset{x\to 1}{lim}5-3x=2$

Hence, continuous at $x=1$

For continuity at $x=-2$

$RHL=\underset{x\to -2}{lim}5-3x=11$

$LHL=\underset{x\to -2}{lim}\frac{6}{x-10}=-\frac{1}{2}$

As $LHL\ne RHL$ it is not continuous at $x=-2$