\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)
(If there's a better way to denote the limit, please let me know; I've only just started playing around with TeX.)
Steps:
\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\) [Root Law]
\(\displaystyle (\sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x)*(1/x)\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9 + 1/x} + \buildrel Lim\over{x\to\infty} - 3\)
\(\displaystyle \sqrt{9 + 0} + - 3\) [Taking the limit]
\(\displaystyle 0\)
The book claims \(\displaystyle 1/6\), but I'm not sure where they'd get that answer.
--Pontifex
(If there's a better way to denote the limit, please let me know; I've only just started playing around with TeX.)
Steps:
\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\) [Root Law]
\(\displaystyle (\sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x)*(1/x)\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9 + 1/x} + \buildrel Lim\over{x\to\infty} - 3\)
\(\displaystyle \sqrt{9 + 0} + - 3\) [Taking the limit]
\(\displaystyle 0\)
The book claims \(\displaystyle 1/6\), but I'm not sure where they'd get that answer.
--Pontifex