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 $\frac{\sqrt{x^2 + 1}}{x}$?
Hint: Observe that, for $x&gt;0$,
$$\large\sqrt{x^2+1\strut}=\sqrt{x^2(1+\tfrac{1}{x^2})}=\sqrt{x^2\strut} \;\cdot\;\sqrt{1+\tfrac{1}{x^2}}=x\;\cdot\;\sqrt{1+\tfrac{1}{x^2}}$$
$$
\frac{\sqrt{x^2 + 1}}{x} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \sqrt{\frac{x^2}{x^2} + \frac 1 {x^2}} = \cdots
$$

$\frac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{\sqrt{x^2 + 1} + x}{x}\right)}$ is exactly the same thing as $\dfrac 1 {\sqrt{x^2+1}+x}$.  To get that just multiply the numerator and the denominator by $x$.  Then:
$$
0 &lt; \frac 1 {\sqrt{x^2+1}+x}&lt;\frac 1 {2x} \to 0 \text{ as } x\to\infty.
$$