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artofspeedBest ResponseYou've already chosen the best response.0
are a, b variables or constant
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
put y=1/x as x>infinity, y>0
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Yup..That Make Sense Thxxx
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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 (pq)(p+q)=p²q²:\[\frac{ x^2+ax+bx^2 }{ \sqrt{x^2+ax+b}+x }=\frac{ ax+b }{\sqrt{x^2+ax+b}+x }\]
 one year ago

RadEnBest ResponseYou've already chosen the best response.0
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 = (bq)/(2sqrt(a))
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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\[\frac{ a+0 }{ \sqrt{1+0+0} +1}=a\]
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Once you see you could use this trick, everything goes (almost) by itself ;)
 one year ago
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