lim limits_{x rightarrow 0} left(tan left(fra | vedclass

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(D) The given limit is of the form $1^{\infty}$.We use the formula $\lim \limits_{x \rightarrow a} f(x)^{g(x)} = e^{\lim \limits_{x \rightarrow a} g(x

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$e^{2}$

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(D) The given limit is of the form $1^{\infty}$.We use the formula $\lim \limits_{x ightarrow a} f(x)^{g(x)} = e^{\lim \limits_{x ightarrow a} g(x)(f(x)-1)}$.Here,$f(x) = 	an(\frac{\pi}{4} + x)$ and $g(x) = \frac{1}{x}$.$\lim \limits_{x ightarrow 0} \left(	an \left(\frac{\pi}{4}+xight)ight)^{\frac{1}{x}} = e^{\lim \limits_{x ightarrow 0} \frac{1}{x} \left(	an \left(\frac{\pi}{4}+xight)-1ight)}$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $	an(\frac{\pi}{4}+x) = \frac{1+	an x}{1-	an x}$.So,$	an(\frac{\pi}{4}+x)-1 = \frac{1+	an x - (1-	an x)}{1-	an x} = \frac{2	an x}{1-	an x}$.Thus,the exponent becomes $\lim \limits_{x ightarrow 0} \frac{2	an x}{x(1-	an x)} = \lim \limits_{x ightarrow 0} 2 \cdot \frac{	an x}{x} \cdot \frac{1}{1-	an x} = 2 \cdot 1 \cdot 1 = 2$.Therefore,the limit is $e^{2}$.

$\lim \limits_{x \rightarrow 0} \left(\tan \left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}}$ is equal to

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