Find the limit as $x$ approaches positive infinity. $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt{16 x^{4}-3}}{2 x^{2}+3}=$

Full solution

Q. Find the limit as

x

approaches positive infinity.

\newline

\lim _{x \rightarrow \infty} \frac{\sqrt{16 x^{4}-3}}{2 x^{2}+3}=

Identify highest power of x: Identify the highest power of x in the numerator and denominator. $\newline$ In the expression $(\sqrt{16x^{4}-3})/(2x^{2}+3)$ , the highest power of x in the numerator inside the square root is $x^4$ , and the highest power of x in the denominator is $x^2$ .
Divide numerator and denominator: Divide the numerator and the denominator by $x^2$ , the highest power of $x$ in the denominator. $\newline$ $\lim_{x \rightarrow \infty}\left(\frac{\sqrt{16x^{4}-3}}{2x^{2}+3}\right) = \lim_{x \rightarrow \infty}\left(\frac{\sqrt{(16x^{4}-3)/x^4}}{2+(3/x^2)}\right)$ $\newline$ $= \lim_{x \rightarrow \infty}\left(\frac{\sqrt{16 - 3/x^4}}{2 + 3/x^2}\right)$
Simplify expression as $x$ approaches infinity: Simplify the expression inside the square root and the denominator as $x$ approaches positive infinity. $\newline$ As $x$ approaches positive infinity, the terms $\frac{3}{x^4}$ and $\frac{3}{x^2}$ approach $0$ . $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt{16 - \frac{3}{x^4}}}{2 + \frac{3}{x^2}}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{16}}{2}\right)$ $\newline$ $= \frac{\sqrt{16}}{2}$
Calculate final value of the limit: Calculate the final value of the limit. $\sqrt{16}/2 = 4/2 = 2$