lim _{x rightarrow 0}left[tan left(x+frac{pi} | vedclass

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(A) Let $L = \lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)ight]^{1 / x}$.Since this is of the form $1^\infty$,we use the formula $\

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$e^2$

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(A) Let $L = \lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)ight]^{1 / x}$.Since this is of the form $1^\infty$,we use the formula $\lim_{x 	o a} [f(x)]^{g(x)} = e^{\lim_{x 	o a} [f(x)-1]g(x)}$.$L = e^{\lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)-1ight] \frac{1}{x}}$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $	an(x+\frac{\pi}{4}) = \frac{	an x + 1}{1 - 	an x}$.$L = e^{\lim _{x ightarrow 0}\left[\frac{	an x+1}{1-	an x}-1ight] \frac{1}{x}} = e^{\lim _{x ightarrow 0} \left[\frac{	an x+1-1+	an x}{1-	an x}ight] \frac{1}{x}}$.$L = e^{\lim _{x ightarrow 0} \frac{2 	an x}{x(1-	an x)}}$.Since $\lim_{x 	o 0} \frac{	an x}{x} = 1$ and $\lim_{x 	o 0} (1-	an x) = 1$,we get $L = e^{2 	imes 1 / 1} = e^2$.

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

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