Find all point of discontinuity of the function

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 $x=\frac{1}{2}, 2$

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