Question

http://img718.imageshack.us/img718/6802/1ac88e33858a4381971c9b1.png

Can anyone show me how you get that answer? I tried all kinds of way to manipulate the limit but I can't seem to find the right way

Answer 1

I assume you already tried multiplying by the conjugate?

Answer 2

multiply it by\[{\sqrt{5x+1}+\sqrt{5x}\over\sqrt{5x+1}+\sqrt{5x}}\]at least that's how these usually work...

Answer 3

I tried and ended up with \[\sqrt(x)+1/(\sqrt(5x+1)+\sqrt(5x))\] not sure where to go from here

Answer 4

let me actually try it

Answer 5

ok that's not quite right...

Answer 6

oh i think there isn't a +1

Answer 7

\[{\sqrt x (\sqrt{5x+1}-\sqrt{5x})\over1}\cdot{\sqrt{5x+1}+\sqrt{5x}\over\sqrt{5x+1}+\sqrt{5x}}={\sqrt x(5x+1-5x)\over\sqrt{5x+1}+\sqrt{5x}}\]\[={\sqrt x\over\sqrt{5x+1}+\sqrt{5x}}\]now divide top and bottom by \(\sqrt x\)

Answer 8

so yes you realized your mistake

Answer 9

do you see the answer yet?

Answer 10

no I'm stuck with sqrt(x) times the denominator :(

Answer 11

\[={\sqrt x\cdot\frac1{\sqrt x}\over(\sqrt{5x+1}+\sqrt{5x})\cdot\frac1{\sqrt x}}={1\over\sqrt{5+\frac1x}+\sqrt5}\]now let x go to infinity

Answer 12

okay now I see

Answer 13

thanks again :)

Answer 14

no prob :)