University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\infty}\frac{x-3}{\sqrt{4x^2+25}}=\frac{1}{2}$$
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 $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to\infty}\frac{x-3}{\sqrt{4x^2+25}}$$ The highest power of $x$ in the denominator here, as you can see, is $x^2$. However, as $x^2$ is wrapped inside a square root, to eliminate $x^2$ in square root, we only need to divide both numerator and denominator by $x$: $$A=\lim_{x\to\infty}\frac{1-\frac{3}{x}}{\frac{\sqrt{4x^2+25}}{x}}$$ In the denominator, 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 $x=\sqrt{x^2}$. $$A=\lim_{x\to\infty}\frac{1-\frac{3}{x}}{\frac{\sqrt{4x^2+25}}{\sqrt{x^2}}}=\lim_{x\to\infty}\frac{1-\frac{3}{x}}{\sqrt{\frac{4x^2+25}{x^2}}}$$ $$A=\lim_{x\to\infty}\frac{1-\frac{3}{x}}{\sqrt{4+\frac{25}{x^2}}}=\frac{\lim_{x\to\infty}1-\lim_{x\to\infty}\frac{3}{x}}{\sqrt{\lim_{x\to\infty}4+\lim_{x\to\infty}\frac{25}{x^2}}}$$ $$A=\frac{1-0}{\sqrt{4+0}}=\frac{1}{\sqrt4}=\frac{1}{2}$$