Answer with Step-by-step explanation:
We are given that
[tex]f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.[/tex]
We have to explain that why the function is discontinuous at x=2
We know that if function is continuous at x=a then LHL=RHL=f(a).
[tex]f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}[/tex]
LHL=Left hand limit when x <2
Substitute x=2-h
where h is small positive value >0
[tex]\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}[/tex]
[tex]\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}[/tex]
Right hand limit =RHL when x> 2
Substitute
x=2+h
[tex]\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}[/tex]
[tex]=\frac{2}{4}=\frac{1}{2}[/tex]
LHL=RHL=[tex]\frac{1}{2}[/tex]
f(2)=1
[tex]LHL=RHL\neq f(2)[/tex]
Hence, function is discontinuous at x=2
