In the third function from this image. The function when $x$ is not equal to $2$ can be transformed into $ x+1 $, resulting in having a value of $3$ at the point $ x = 2 $.
Why is this not considered a removable discontinuity?
2026-03-27 00:09:54.1774570194
Why this piecewise function doesn't have a removable discontinuity?
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It is because
$f(x)=\begin{cases}\frac{x^2-x-2}{x-2}; x\ne 2\\ 3; x=2\end{cases}$
is continuous at $x=2$ (Note that $\lim _{x\to 2} f(x)=\lim_{x\to 2}(x+1)=3$ and $f(2)=3$) and as per the given question, you are required to choose functions which have a removable discontinuity at $x=2$.