Could someone show the steps in working out this limit:
limit (k^2*sqrt(9k^2-4k+1))/(3k^3+k-7) as k goes from 3 to infinity
thanks
$\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{{{k^2}\sqrt {9{k^2} - 4k + 1} }}{{3{k^3} + k - 7}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{k^3}\sqrt {9 - \frac{4}
{k} + \frac{1}
{{{k^2}}}} }}
{{{k^3}\left( {3 + \frac{1}
{{{k^2}}} - \frac{7}
{{{k^3}}}} \right)}} =$
$\displaystyle = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {9 - \frac{4}
{k} + \frac{1}
{{{k^2}}}} }}
{{3 + \frac{1}
{{{k^2}}} - \frac{7}
{{{k^3}}}}} = \frac{{\sqrt {9 - 0 + 0} }}
{{3 + 0 - 0}} = \frac{3}
{3} = 1.$