On rationalising numerator, we get
=limx→0(√1+3x−√1−3x)(√1+3x+√1−3x)x(√1+3x+√1−3x)
=limx→0((√1+3x)2−(√1−3x)2)x(√1+3x+√1−3x)
=limx→0(1+3x−1+3x)x(√1+3x+√1−3x)
=limx→06xx(√1+3x+√1−3x)
=limx→06(√1+3x+√1−3x) [∵x≠0]
=6x(√1+3(0)+√1−3(0))
=62=3
∴limx→0√1+3x−√1−3xx=3