A community for students.
Here's the question you clicked on:
 0 viewing
Yahoo!
 2 years ago
\[\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b}  x ) = ?\]
Yahoo!
 2 years ago
\[\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b}  x ) = ?\]

This Question is Closed

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 ;)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.