Question

\[\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b} - x ) = ?\]

Answer 1

lolwut

Answer 2

are a, b variables or constant

Answer 3

Constant

Answer 4

put y=1/x
as x->infinity, y->0

Answer 5

ye hartnn is right

Answer 6

then you can apply LH.

Answer 7

Yup..That Make Sense Thxxx

Answer 8

welcome ^_^

Answer 9

xxx

Answer 10

It can also be done without l'Hopital.
\[\sqrt{x^2+ax+b}-x=(\sqrt{x^2+ax+b}-x) \cdot \frac{ \sqrt{x^2+ax+b}+x }{\sqrt{x^2+ax+b}+x }=\]using (p-q)(p+q)=p²-q²:\[\frac{ x^2+ax+b-x^2 }{ \sqrt{x^2+ax+b}+x }=\frac{ ax+b }{\sqrt{x^2+ax+b}+x }\]

Answer 11

Lol..This Also..Helps

Answer 12

alternative :) use the formula :
if given : lim (x->~) sqrt(ax^2+bx+c) - sqrt(px^2+qx+r)
with a=p, then the limit value's is
L = (b-q)/(2sqrt(a))

Answer 13

It is not yet ready...
Divide everything by x:\[\frac{ a+\frac{ b }{ x } }{ \frac{ \sqrt{x^2+ax+b} }{ x }+1 }=\frac{ a+\frac{ b }{ x } }{ \sqrt{\frac{ x^2+ax+b }{ x^2 }} +1}=\frac{ a+\frac{ b }{ x } }{ \sqrt{1+\frac{ a }{ x }+\frac{ b }{ x^2 }} +1}\]
Now if you let x go to infinity, you get\[\frac{ a+0 }{ \sqrt{1+0+0} +1}=a\]

Answer 14

1+1 = 2, ZeHanz

Answer 15

Once you see you could use this trick, everything goes (almost) by itself ;)