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f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),i...

`f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),if -1 le x lt 0),((2x+1)/(x-1),if 0 le x le 1):}` at `x = 0`.

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The correct Answer is:
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We have, `f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),if -1 le x lt 0),((2x+1)/(x-1),if 0 le x le 1):}` at `x = 0`.
`:. LHL = underset(xrarr0^(-))(lim)(sqrt(1+kx)-sqrt(1-kx))/(x)`
`= underset(xrarr0^(-))(lim)((sqrt(1+kx)-sqrt(1-kx))/(x)). ((sqrt(1+kx)-sqrt(1+kx))/(sqrt(1+kx)-sqrt(1+kx)))`
`= underset(xrarr0^(-))(lim)(1+kx-1+kx)/(x[sqrt(1+kx)+sqrt(1-kx)])`
`= underset(xrarr0^(-))lim(2kx)/(xsqrt(1+kx)+sqrt(1-kx))`
`= underset(hrarr0)lim(2k)/(sqrt(1+k(0-h))+sqrt(1-k(0-h)))`
` = underset(hrarr0)(lim)(2k)/(sqrt(1-kh) +sqrt(1+kh)) = (2k)/(2) = k`
and `f(0) = (2xx 0+1)/(0-1) = - 1`
`rArr k = - [:' LHL = RHL = f(0)]`
`rArr k = -1`
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