Question:
Find all point of discontinuity of the function $f(t)=\frac{1}{t^{2}+t-2}$, where $t=\frac{1}{x-1}$
Solution:
$f(t)=\frac{1}{t^{2}+t-2}$
Now, let $u=\frac{1}{x-1}$
$\therefore f(u)=\frac{1}{u^{2}+2 u-u-2}=\frac{1}{u^{2}+u-2}=\frac{1}{(u+2)(u-1)}$
So, $f(u)$ is not defined at $u=-2$ and $u=1$
If $u=-2$, then
$-2=\frac{1}{x-1}$
$\Rightarrow 2 x=1$
$\Rightarrow x=\frac{1}{2}$
If $u=1$, then
$1=\frac{1}{x-1}$
$\Rightarrow x=2$
Hence, the function is discontinuous at $x=\frac{1}{2}, 2$