1. Let $L=\lim_{x\to 0}\left(\tan\left(\frac\pi4+x\right)\right)^{1/x}.$ This is of the form $1^{\infty}$, so we take logarithm.

2. Put $\ln L=\lim_{x\to 0}\frac{1}{x}\ln\left(\tan\left(\frac\pi4+x\right)\right).$ So we need $\ln L=\lim_{x\to 0}\frac{\ln\left(\tan\left(\frac\pi4+x\right)\right)-\ln(1)}{x}.$ Since $\tan(\pi/4)=1$, this is the derivative of $f(x)=\ln\left(\tan\left(\frac\pi4+x\right)\right)$ at $x=0$.

3. Differentiate: $f'(x)=\frac{1}{\tan\left(\frac\pi4+x\right)}\cdot \sec^2\left(\frac\pi4+x\right).$ Hence f'(0)=\frac{\sec^2(\pi/4)}{\tan(\pi/4)}=rac{2}{1}=2. Therefore, $\ln L=2.$

4. Exponentiating, $L=e^2.$

5. Checking options:

• A: $2$ ❌
• B: $1$ ❌
• C: $e$ ❌
• D: $e^2$ ✅

So the correct answer is D.