Evaluate LaTeX: \displaystyle \lim_{x \to \infty} \left(1 + \frac{8}{x}\right)^{\frac{3 x}{2}}

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{8}{x}\right)^{\frac{3 x}{2}} \right) Pulling the limit out of the continuous function and using log properties gives: LaTeX: \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{8}{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{8}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{8}{x^{2}}}{- \frac{2}{3 x^{2}}} = 12 Solving for LaTeX: \displaystyle L gives LaTeX: \displaystyle L = e^{12}