Question

evaluate the lim as x approaches infinity: sqrt(x^2+2x)-sqrt(x^2-x).
wolframalpha is giving me 3/2 but i keep getting 1. (inf/inf). i know wolfram is right but i dont know why

Answer 1

multiply top and bottom by
\[\sqrt{x^2+2x}+\sqrt{x^2-x}\]

Answer 2

btw
\[\frac{\infty}{\infty}\neq 1\] in general
it is a statement about limits. if you get
\[\frac{\infty}{\infty}\] i means you have to do more work

Answer 3

oh alright i did not know that. so i did that first step and got 3x/(sqrt(x^2+2x)+sqrt(x^2-x))

Answer 4

thats where i got inf/inf. so i dont know what to do from this point

Answer 5

oh ok

Answer 6

best thing to do is reason it out
\[\sqrt{x^2}=x\] so it is like having a denominator of
\[2x\] giving you
\[\frac{3x}{2x}=\frac{3}{2}\]

Answer 7

if that is unsatisfactory, divide top and bottom by
\[x\] to get
\[\frac{3}{\sqrt{1+\frac{2}{x}}+\sqrt{1-\frac{1}{x}}}\] then you can see the 3/2

Answer 8

ok awesome
tyvm