Question

lim x-> infinity ((sqrt(x^2-3x)-x) how do i find the limit? please help don't understand

Answer 1

try multiplying by
\[\frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]

Answer 2

i tried that gives me -3x/sqrt(x^2-3x)+x dont know what to do after

Answer 3

that is -infinity/infinity+infinity

Answer 4

but it needs to be a number right?

Answer 5

you get
\[\frac{-3x}{\sqrt{x^2-3x}+x}\] yes

Answer 6

now you can pretty much do this by your eyeballs.
ignore the \(-3x\) inside the radical in the denominator
you get \[\frac{-3x}{\sqrt{x^2}+x}\]
\[=\frac{-3x}{x+x}=\frac{-3x}{2x}=-\frac{3}{2}\]

Answer 7

oh yeah thats right i understand tnx

Answer 8

if that is not sufficient, you have to divide top and bottom by \(x\)
be careful inside the radical

Answer 9

yw