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

Mathematics
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lolwut
are a, b variables or constant
Constant

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Other answers:

put y=1/x as x->infinity, y->0
ye hartnn is right
then you can apply LH.
Yup..That Make Sense Thxxx
welcome ^_^
xxx
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:\[\frac{ x^2+ax+b-x^2 }{ \sqrt{x^2+ax+b}+x }=\frac{ ax+b }{\sqrt{x^2+ax+b}+x }\]
Lol..This Also..Helps
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))
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\[\frac{ a+0 }{ \sqrt{1+0+0} +1}=a\]
1+1 = 2, ZeHanz
Once you see you could use this trick, everything goes (almost) by itself ;)

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