Question

Help! Limit as x approaches infinity of (sqrt(x^2-x)-x)?

Answer 1

\[[\lim_{x \rightarrow \infty} \left( \sqrt{x^2-x}-x \right)]\]

Answer 2

-1

Answer 3

Is this true? and if so, how?

Answer 4

because when you graph it , it shows y is at -1

Answer 5

Are you able to proof it mathimatically without a graph? That's what I need.

Answer 6

It is -1/2=-.5

Answer 7

can someone walk me through the steps?

Answer 8

\[
\frac {\left( \sqrt{x^2-x}-x \right)\left( \sqrt{x^2-x}+x \right)}
{\left( \sqrt{x^2-x}+x \right)}=\\
\frac{x^2-x-x^2}{\left( \sqrt{x^2-x}+x \right)}=\frac {-x}{\left( \sqrt{x^2-x}+x \right)}
\]
The last expression goes to -1/2 when x goes to Infinity

Answer 9

ok up to here. I understand now. Now, how did you get -1/2 when it goes to infinity?

Answer 10

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

Answer 11

For the last equality divide up and down by x

Answer 12

this guy is 2 smart

Answer 13

\[
\frac {-1}{\left( \sqrt{1-\frac 1 x}+1 \right)}
\]
goes to -1/2 when x goes to Infinity

Answer 14

Did you get it?

Answer 15

wait eliassaab when you did that multiplying did you multiply by the conjuant? or however its spelled

Answer 16

yes, I think so. IN other words when the -1/x goes to infinity, it is such a negligible difference that we don't count it. It all makes sense now. You are a life savor!

Answer 17

yes

Answer 18

do you know how to solve functions that have a floor function in them as well?

Answer 19

Close this problem and ask the new one with floor function.

Answer 20

ok