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artofspeed
 2 years ago
Best ResponseYou've already chosen the best response.0are a, b variables or constant

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1put y=1/x as x>infinity, y>0

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

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1It 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:\[\frac{ x^2+ax+bx^2 }{ \sqrt{x^2+ax+b}+x }=\frac{ ax+b }{\sqrt{x^2+ax+b}+x }\]

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0alternative :) 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))

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1It 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\[\frac{ a+0 }{ \sqrt{1+0+0} +1}=a\]

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Once you see you could use this trick, everything goes (almost) by itself ;)
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