## 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 $x\lt0$, so $x=-\sqrt{x^2}$. $$A=\lim_{x\to-\infty}\frac{\frac{\sqrt{x^2+1}}{-\sqrt{x^2}}}{1+\frac{1}{x}}=\lim_{x\to-\infty}\frac{-\sqrt{\frac{x^2+1}{x^2}}}{1+\frac{1}{x}}$$ $$A=-\lim_{x\to-\infty}\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}=-\frac{\sqrt{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{1}{x^2}}}{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{1}{x}}$$ $$A=-\frac{\sqrt{1+0}}{1+0}=-\frac{\sqrt1}{1}=-1$$