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If lim(x rarr 0)[1+x ln(1+b^2)]^(1/x)=2b...

If `lim_(x rarr 0)[1+x ln(1+b^2)]^(1/x)=2b sin^2 theta, b gt 0 "and" theta in (-pi, pi]`. then the value of `theta` is

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The correct Answer is:
D
`e^(ln(1+b^(2)))=2bsin^(2)theta`
or `" "sin^(2)theta=(1+b^(2))/(2b)=1" as "(1+b^(2))/(2b)ge1`
`theta=+-pi//2`
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