lim (x^2+6x+1)^(1/2)-x
x -> +inf
I know the answer is 3, but I don't know how to find that without using a table.
$\displaystyle \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 6x + 1} - x} \right) = \frac{{\left( {\sqrt {{x^2} + 6x + 1} - x} \right)\left( {\sqrt {{x^2} + 6x + 1} + x} \right)}}{{\left( {\sqrt {{x^2} + 6x + 1} + x} \right)}} =$
$\displaystyle = \mathop {\lim }\limits_{x \to + \infty } \frac{{6x + 1}}
{{\sqrt {{x^2} + 6x + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {6 + \frac{1}{x}} \right)}}{{x\left( {\sqrt {1 + \frac{6}{x} + \frac{1}{{{x^2}}}} + 1} \right)}} =$
$\displaystyle = \mathop {\lim }\limits_{x \to + \infty } \frac{{6 + \frac{1}
{x}}}
{{\sqrt {1 + \frac{6}
{x} + \frac{1}
{{{x^2}}}} + 1}} = \frac{{6 + 0}}
{{\sqrt {1 + 0 + 0} + 1}} = \frac{6}
{2} = 3.$