Evaluate $LaTeX: \displaystyle \lim_{x \to \infty} \left(1 + \frac{9}{x}\right)^{\frac{2 x}{7}}$

This is an indeterminate form of the type $LaTeX: \displaystyle 1^\infty$ . Taking the natural logarithm of both sides gives: $LaTeX: \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{9}{x}\right)^{\frac{2 x}{7}} \right)$ Pulling the limit out of the continuous function and using log properties gives: $LaTeX: \ln(L) = \lim_{x \to \infty}\frac{2 x}{7}\ln\left(1 + \frac{9}{x} \right)$ This is an indeterminate form of the type $LaTeX: \displaystyle 0 \cdot \infty$ . Converting it to type $LaTeX: \displaystyle \frac{0}{0}$ and using L'Hospitials rule gives: $LaTeX: \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{9}{x}\right)}{\frac{7}{2 x}} = \frac{- \frac{9}{x^{2}}}{- \frac{7}{2 x^{2}}} = \frac{18}{7}$ Solving for $LaTeX: \displaystyle L$ gives $LaTeX: \displaystyle L = e^{\frac{18}{7}}$