According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does 
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to 
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\sqrt{1 + \dfrac{1}{x^2}} + 1}$$
I understand how he simplifies $\frac{x}{x}$ to 1 but how does he simplify $\frac{\sqrt{x^2 + 1}}{x}$?

Since you are calculating the limit at $x\to\infty$ you may assume that $x>0$. Then
$$\frac{\sqrt{x^2+1}+x}x=\frac{\sqrt{x^2+1}}{\sqrt{x^2}}+\frac xx=\sqrt{\frac{x^2+1}{x^2}}+1=\sqrt{1+\frac{1}{x^2}}+1$$
