Prove that the function f defined by `f(x) = {{:((x)/(|x|+2x^(2)), if x ne 0),(k, if x = 0):}` remains discontinuous at `x = 0`, regardings the choice iof k.

Answer

N/a

Explanation

We have, `f(x) = {{:((x)/(|x|+2x^(2)), if x ne 0),(k, if x = 0):}`
At `x= 0, LHL =underset(xrarr0^(-))(lim)(x)/(|x|+2x^(2)) = underset(hrarr0)(lim)((0-h))/(|0-h|+2(0-h)^(2))`
`= underset(hrarr0)(lim)(-h)/(h+2h^(2))=underset(hrarr0)(lim)(-h)/(h(1+2h)) = -1`
`RHL= underset(xrarr0^(+))lim(x)/(|x|+2x^(2))= underset(hrarr0)(lim)(0+h)/(|0+h|
+2(0+h)^(2))`
`= underset(hrarr0)(lim)(h)/(h+2h)^(2)=underset(hrarr0)(lim)(h)/(h(1+2h)) = 1`
and `f(0) = k`
Since, ` LHL ne RHL ` for any value of k.
Hence, `f(x)` is discountinuous at `x = 0` regardiess the choice of k.