Find the value of k for which f(x) is continuous at x=0, where f(x) is defined as:

f(x) = { √(1+kx) - √(1-kx) / x if -1 ≤ x < 0
{ 2x+1 if 0 ≤ x < 1

The given function f(x) is continuous at x = 0.

limx→0+ f(x) = limx→0+ (2x+1) = 1

limx→0- f(x) = limx→0- (√(1+kx) - √(1-kx) / x)

To evaluate this limit, we can use L'Hôpital's rule or rationalize the numerator:

limx→0- (√(1+kx) - √(1-kx) / x) = limx→0- [(√(1+kx) - √(1-kx)) * (√(1+kx) + √(1-kx)) / x(√(1+kx) + √(1-kx))]

= limx→0- [(1+kx) - (1-kx) / x(√(1+kx) + √(1-kx))]

= limx→0- [2kx / x(√(1+kx) + √(1-kx))]

= limx→0- [2k / (√(1+kx) + √(1-kx))]

= 2k / (√1 + √1) = 2k / 2 = k

For f(x) to be continuous at x=0, we must have:

limx→0- f(x) = limx→0+ f(x) = f(0)

k = 1

Therefore, the value of k is 1.