Q. limx→0(1−12x)(1√tanx+4−2)=
Solution:
G.E=limx→0(2x−12x)(√tanx+4+2tanx+4−4)
=limx→0(2x−1x)xtanx(√tanx+4+22x)
=log2.1.4=4log2=log16
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