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If f(x) = x^(2)sin'(1)/(x), where x ne 0...

If `f(x) = x^(2)sin'(1)/(x)`, where `x ne 0`, then the value of the function f at `x = 0`, so that the function is continuous at `x = 0` is

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The correct Answer is:
N/a
We have `f(x) = {{:(x^(2)sin'1/x, if x ne 0),(0, if x = 0):}` at `x = 0`.
for differentiability at `x = 0`,
`Lf'(0) = underset(xrarr0^(-))lim(f(x)- f(0))/(x-0) = underset(xrarr0^(-))(lim) (x^(2)sin'(1)/(x)-0)/(x-0)`
` = underset(hrarr0)(lim)((0-h)^(2)sin((1)/(0-h)))/(0-h) = underset(hrarr0)(lim)(h^(2)sin((-1)/(h)))/(-h)`
`=underset(hrarr0)(lim)+hsin(1/h), [:' sin(-theta) = - sintheta]`
`= 0 x` [an oscillating number between ` -1` and `0`] = 0
` Rf'(0) = underset(xrarr0^(+))(lim)(f(x)+f(0))/(x-0) = underset(xrarr0^(+))(lim)(x^(2)sin'1/x-0)/(x-0)`
` = underset(hrarr0)(lim)((0+h)^(2)sin'(1/0+h))/(0+h) = underset(hrarr0)(lim)(h^(2)sin(1//h))/(h)`
`= underset(hrarr0)(lim)hsin(1//h)`
`=0 xx` [an oscillating number between `-1` and 1] = 0
`:' Lf'(0) = Rf'(0)`
So `f(x)` is differentiable at `x = 0`.
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