Calculus: Limits, Continuity & Differentiability

This is an indeterminate form of type $1^\infty$. Taking natural logarithm of both sides: $\ln L = \lim_{x \to \infty} x \ln\left(1 + \frac{6}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{6}{x}\right)}{\frac{1}{x}}$ Applying L'Hospital's rule: $\ln L = \lim_{x \to \infty} \frac{\frac{1}{1 + k/x} \cdot \left(-\frac{k}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{6}{1 + k/x} = 6$ Exponentiating both sides: $L = e^{6} = 403.43$