If lim _{x rightarrow 0}left{1+x log left(1+a | vedclass

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(A) Given $\lim _{x ightarrow 0}\left\{1+x \log \left(1+a^2ight)ight\}^{1 / x} = 2 a \sin ^2 	heta$.Using the standard limit $\lim _{x ightar

Vedclass · Accepted Answer

$	heta = n \pi + \frac{\pi}{2}, (n \in Z)$

Vedclass · Answer

(A) Given $\lim _{x ightarrow 0}\left\{1+x \log \left(1+a^2ight)ight\}^{1 / x} = 2 a \sin ^2 	heta$.Using the standard limit $\lim _{x ightarrow 0} (1+f(x))^{1/x} = e^{\lim _{x ightarrow 0} \frac{f(x)}{x}}$, we get:$e^{\lim _{x ightarrow 0} \frac{x \log(1+a^2)}{x}} = e^{\log(1+a^2)} = 1+a^2$.Equating this to the given expression:$1+a^2 = 2a \sin^2 	heta$.Rearranging as a quadratic in $a$:$a^2 - (2 \sin^2 	heta)a + 1 = 0$.For $a$ to be a real number, the discriminant $D \ge 0$:$D = (2 \sin^2 	heta)^2 - 4(1)(1) \ge 0$$\Rightarrow 4 \sin^4 	heta - 4 \ge 0$$\Rightarrow \sin^4 	heta \ge 1$.Since $\sin^4 	heta \le 1$ for all $	heta$, the only possibility is $\sin^4 	heta = 1$.$\Rightarrow \sin^2 	heta = 1 = \sin^2 \frac{\pi}{2}$.$\Rightarrow 	heta = n\pi \pm \frac{\pi}{2}$, which simplifies to $	heta = n\pi + \frac{\pi}{2}$ for $n \in Z$.

If $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}=2 a \sin ^2 \theta$, where $a>0$ and $\theta \in R$, then:

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