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\[\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b}  x ) = ?\]
 one year ago
 one year ago
Yahoo! Group Title
\[\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b}  x ) = ?\]
 one year ago
 one year ago

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

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

artofspeed Group TitleBest ResponseYou've already chosen the best response.0
ye hartnn is right
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
then you can apply LH.
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago

ZeHanz Group TitleBest 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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Lol..This Also..Helps
 one year ago

RadEn Group TitleBest 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

ZeHanz Group TitleBest 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

RadEn Group TitleBest ResponseYou've already chosen the best response.0
1+1 = 2, ZeHanz
 one year ago

ZeHanz Group TitleBest 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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