Limit Exponent 8Bf3E8

1. **Problem statement:** Explain why $\lim_{x\to\infty}x^{1/x}=1$. 2. **Use logarithms to simplify the exponent.** Let $y=x^{1/x}.$ Taking the natural log gives $\ln y=\frac{\ln x}{x}.$ So it is enough to find $\lim_{x\to\infty}\frac{\ln x}{x}$. 3. **Evaluate the simpler limit.** As $x\to\infty$, $\ln x$ grows much more slowly than $x$. Using L'Hôpital's rule: $$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac{1}{x}=0.$$ 4. **Go back to $x^{1/x}$.** Since $\ln y\to 0$, we get $$y=e^{\ln y}\to e^0=1.$$ 5. **Final answer:** $\lim_{x\to\infty}x^{1/x}=1$ because its logarithm goes to $0$, and $e^0=1$.