Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (b)/Solution 1

Rationalizing the numerator gives

${\begin{aligned}&\lim _{x\to \infty }{\sqrt {9x^{2}+x}}-3x\\=&\lim _{x\to \infty }({\sqrt {9x^{2}+x}}-3x)\cdot {\frac {{\sqrt {9x^{2}+x}}+3x}{{\sqrt {9x^{2}+x}}+3x}}\\=&\lim _{x\to \infty }{\frac {9x^{2}+x-9x^{2}}{{\sqrt {9x^{2}+x}}+3x}}\\=&\lim _{x\to \infty }{\frac {x}{{\sqrt {9x^{2}+x}}+3x}}\end{aligned}}$

Now we can factor the denominator. Remembering that ${\sqrt {x^{2}}}=|x|$, we have that

${\begin{aligned}&\lim _{x\to \infty }{\frac {x}{{\sqrt {9x^{2}+x}}+3x}}\\=&\lim _{x\to \infty }{\frac {x}{|x|{\sqrt {9+1/x}}+3x}}\end{aligned}}$

As the limit goes to positive infinity, $|x|=x$ and so

${\begin{aligned}&\lim _{x\to \infty }{\frac {x}{|x|{\sqrt {9+1/x}}+3x}}\\=&\lim _{x\to \infty }{\frac {x}{x{\sqrt {9+1/x}}+3x}}\\=&\lim _{x\to \infty }{\frac {x}{x({\sqrt {9+1/x}}+3)}}\\=&\lim _{x\to \infty }{\frac {1}{{\sqrt {9+1/x}}+3}}\\=&{\frac {1}{{\sqrt {9+0}}+3}}\\=&{\frac {1}{6}}\\\end{aligned}}$