Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 317: 63

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$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} \cr & {\text{Evaluating the limit when }}x \to \infty \cr & \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = {x^{1/\infty }} = {\infty ^0} \cr & {\text{This limit has the form }}{\infty ^0}{\text{ }} \cr & {x^{1/x}} = {e^{\frac{1}{x}\ln x}},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln x}} \cr & {\text{The first step is to evaluate }} \cr & L = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln x = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{x} = \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln x} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} \cr & L = 0 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to \infty } {x^{1/x}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^0} = {e^L} = 1 \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 + x} \right)^{1/x}} = {e^L} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {1 + x} \right)^{1/x}} = 1 \cr} $$