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f(x) = {{:((2x^(2)-3x-2)/(x-2), if x ne ...

`f(x) = {{:((2x^(2)-3x-2)/(x-2), if x ne 2), (5, if x = 2):}` at `x = 2`.

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We have , `f(x) = {{:((2x^(2)-3x-2)/(x-2), if x ne 2), (5, if x = 2):}` at `x = 2`.
At `x = 2, LHL = underset(xrarr2^(-))lim(2x^(2) - 3x - 2)/(x-2)`
` =underset(hrarr0)(lim)(2-(2-h)^(2)-3(2-h)-2)/((2-h)-2)`
`=underset(hrarr0)(lim)(8+2h^(2)-8h - 6 + 3h - 2)/(-h)`
`=underset(hrarr0)(lim)(2h^(2)-5h)/(-h) = underset(hrarr0)(lim)(h(2h-5))/(-h) = 5`
`RHL = underset(xrarr2^(+))lim(2x^(2)-3x-2)/(x-2)`
`= underset(hrarr0)lim(2(2+h)^(2)-3(2+h)^(2)-2)/((2+h)-2)`
`=underset(hrarr0)(lim)(8+2h^(2)+8h-6-3h-2)/(h)`
`=underset(hrarr0)(lim)(2h^(2)+5h)/(h) = underset(hrarr0)(lim)(2h^(2)+5h)/(h) = underset(hrarr0)(lim)(h(2h+5))/(h) = 5`
and `f(2) = 5`
`:. LHL = RHL = f(2)`
So, `f(x)` is continous at `x = 2`.
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