Question

Evaluate:
limit as x--> infty of (sqrt{x^2+9x+1}-x)

Answer 1

\[\sqrt{x^2+9x+1} -x\] ?

Answer 2

is that what you mean

Answer 3

\[\frac{ (\sqrt{x^2+9x+1} -x) (\sqrt{x^2+9x+1} +x)}{\sqrt{x^2+9x+1} +x}\]

Answer 4

\[= \frac{ (x^2+9x+1)-(x)^2}{\sqrt{ x^2(1 +\frac{9}{x} + \frac{1}{x^2} )} +x}\]

Answer 5

Ok, so just multiply by the conjugate then combine?

Answer 6

pretty much.

It's a standard procedure for anything of that form.

Answer 7

\[= \frac{9x+1}{x [ \sqrt{1+\frac{9}{x}+\frac{1}{x^2}} +1] }\]

Answer 8

\[= \frac{ x (9 + \frac{1}{x} ) }{x[\sqrt{1+\frac{9}{x}+\frac{1}{x^2} } +1} = \frac{9}{2}\]

Answer 9

That last part should say "limit ____" = (9/2) , but I couldn't be bothered writting it.

Answer 10

Just incase you don't get the second last step. Remember lim x-> infinity (1/x^n) , where n>=1 equals zero

Answer 11

so you get \[\frac{9 + (0)}{ \sqrt{1+(0)+(0)} +1} = \frac{9}{2}\]