## Yahoo! 2 years ago $\lim_{x \rightarrow \infty} (\sqrt{x^2 + ax +b} - x ) = ?$

1. artofspeed

lolwut

2. artofspeed

are a, b variables or constant

3. Yahoo!

Constant

4. hartnn

put y=1/x as x->infinity, y->0

5. artofspeed

ye hartnn is right

6. hartnn

then you can apply LH.

7. Yahoo!

Yup..That Make Sense Thxxx

8. hartnn

welcome ^_^

9. artofspeed

xxx

10. ZeHanz

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:$\frac{ x^2+ax+b-x^2 }{ \sqrt{x^2+ax+b}+x }=\frac{ ax+b }{\sqrt{x^2+ax+b}+x }$

11. Yahoo!

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))

13. ZeHanz

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

15. ZeHanz

Once you see you could use this trick, everything goes (almost) by itself ;)