Evaluate the limit

**Scary_Joe** · October 1st 2009, 04:08 PM

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.

**DeMath** · October 1st 2009, 04:16 PM

Originally Posted by Scary_Joe

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.

$\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 6x + 1} - x} \right) = \frac{{\left( {\sqrt {{x^2} + 6x + 1} - x} \right)\left( {\sqrt {{x^2} + 6x + 1} + x} \right)}}{{\left( {\sqrt {{x^2} + 6x + 1} + x} \right)}} =$

$= \mathop {\lim }\limits_{x \to + \infty } \frac{{6x + 1}} {{\sqrt {{x^2} + 6x + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {6 + \frac{1}{x}} \right)}}{{x\left( {\sqrt {1 + \frac{6}{x} + \frac{1}{{{x^2}}}} + 1} \right)}} =$

$= \mathop {\lim }\limits_{x \to + \infty } \frac{{6 + \frac{1} {x}}} {{\sqrt {1 + \frac{6} {x} + \frac{1} {{{x^2}}}} + 1}} = \frac{{6 + 0}} {{\sqrt {1 + 0 + 0} + 1}} = \frac{6} {2} = 3.$

**Scary_Joe** · October 1st 2009, 04:23 PM

Wow, how did you get so good at evaluating limits? Because, when I took calculus I was never taught some of the techniques you used.

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