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f(x)={{:(3x+5, if x ge 2),(x^(3), if x...

`f(x)={{:(3x+5, if x ge 2),(x^(3), if x le 2):}`at `x = 2`

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We have, `f(x)={{:(3x+5, if x ge 2),(x^(3), if x le 2):}`at `x = 2`
At `x = 2 LHL = underset(xrarr2^(-))(lim)(x)^(2)`
`= underset(hrarr0)(lim)(2-h)^(2)=underset(hrarr0)lim(4+h^(2)-4h)=4`
and `RHL = underset(xrarr2^(+))(lim)(3x+5)`
`= underset(hrarr0)(lim)[3(2+h)+5]=11`
Since, ` LHL ne RHL` at `x= 2`
So, `f(x)` is dicontinuous at `x = 2`.
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