Find \( \lim_{{x \to \infty}} (1 + x)^{1/x} \):
- Rewrite with \( e \): \( \lim_{{x \to \infty}} e^{\ln((1+x)^{1/x})} \)
- Put limit in exponent because \( e^u \) is continuous: \( e^{\lim_{{x \to \infty}} (\ln(1+x)/x)} \)
- Now work on the limit:
- The limit \( \lim_{{x \to \infty}} \frac{\ln(1+x)}{x} \) has the form \( \frac{\infty}{\infty} \).
- Apply L’Hôpital’s Rule: \( \lim_{{x \to \infty}} \frac{1/(1+x)}{1} \).
- Simplify: \( \lim_{{x \to \infty}} \frac{1}{1+x} \).
- Take the limit: \( \frac{1}{1+\infty} = 0 \).
- So we get \( e^0 = 1 \).