<s>[INST] 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 $\frac{\sqrt{x^2 + 1}}{x}$?
 [/INST] Hint: Observe that, for $x>0$,
$$\large\sqrt{x^2+1\strut}=\sqrt{x^2(1+\tfrac{1}{x^2})}=\sqrt{x^2\strut} \;\cdot\;\sqrt{1+\tfrac{1}{x^2}}=x\;\cdot\;\sqrt{1+\tfrac{1}{x^2}}$$
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