## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x+1}=1$$
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator. Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x+1}$$ The highest power of $x$ in the denominator here is $x$, so we divide both numerator and denominator by $x$: $$A=\lim_{x\to\infty}\frac{\frac{\sqrt{x^2+1}}{x}}{1+\frac{1}{x}}$$ In the numerator, to put $x$ inside the square root, remember that $|x|=\sqrt{x^2}$ Here since $x\to\infty$, we consider only values of $x\gt0$, so $x=\sqrt{x^2}$. $$A=\lim_{x\to\infty}\frac{\sqrt{\frac{x^2+1}{x^2}}}{1+\frac{1}{x}}=\lim_{x\to\infty}\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x^2}}}{\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x}}$$ $$A=\frac{\sqrt{1+0}}{1+0}=\frac{\sqrt1}{1}=1$$