Question

Limit of (x approaches infinity) of (sqrt (x^2 -3x+4) -x)

Answer 1

Please can anyone help me out?

Answer 2

\[\sqrt{(x^{2} -3x+4) -x}\]
like this?

Answer 3

No the -X is seperate. Limit(x>infinity) of (sqrt x^2 -3x+4) -x)

Answer 4

\[\lim_{x \rightarrow \infty} \sqrt{x^{2}-3x+4}-x\]
so question is, how big is square term compared to -x

Answer 5

I dont know but the answer on the back of the book is -3/2

Answer 6

do i have to divide everything by x^2?

Answer 7

Multiply top and bottom by the conjugate?

\[\frac{ \sqrt{x^{2}-3x+4} - x }{ 1 }*\frac{ \sqrt{x^{2}-3x+4} + x }{ \sqrt{x^{2}-3x+4} + x }\]
Then try and see what you can do.

Answer 8

then i divide it by x which leaves me by -3/2. Thanks a lot!!!! :)

Answer 9

Yep yep :3