Thread: Lim x^{1/x}, x-> infty View Single Post
 P: 5 The standard method of solving indeterminate forms is to apply a transformation to get $$\displaystyle \frac{0}{0}$$ or $$\displaystyle \frac{\infty}{\infty}$$. In this case $$\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left(x^{\frac{1}{x}}\right)}}$$ $$\displaystyle = \lim_{x \to \infty}e^{\frac{1}{x}\ln{x}}$$ $$\displaystyle = \lim_{x \to \infty}e^{\frac{\ln{x}}{x}}$$ $$\displaystyle = e^{\lim_{x \to \infty}\frac{\ln{x}}{x}}$$ from the continutiy of the exponential function, this also gives the required form $$\displaystyle \frac{\infty}{\infty}$$ so that you can use L'Hospital's Rule $$\displaystyle = e^{\lim_{x \to \infty}\frac{\frac{1}{x}}{1}}$$ by L'Hospital's Rule $$\displaystyle = e^{\lim_{x \to \infty}\frac{1}{x}}$$ $$\displaystyle = e^0$$ $$\displaystyle = 1$$.