Question

\[\lim_{x \rightarrow \infty} \sqrt{x^2-x+1}-ax-b\] , a and b are positive constants

Answer 1

@aaronq ?

Answer 2

@ganeshie8 ?

Answer 3

you can break this up...

Answer 4

how?

Answer 5

\[\lim_{x \rightarrow \infty}\sqrt{x^2-x+1}-ax-b=\lim_{x \rightarrow \infty}\sqrt{x^2-x+1}-\lim_{x \rightarrow \infty}ax-\lim_{x \rightarrow \infty}b\]

Answer 6

ok

Answer 7

well, that may not do what you need... let me think for a sec.

Answer 8

rationalizing the numerator may help

Answer 9

\[\lim_{x \rightarrow \infty} \dfrac{ x^2-x+1-(ax+b)^2 }{ \sqrt{x^2-x+1}+ax+b} \]

Answer 10

you could do this...
\[\sqrt{x^2-x+1}-(ax+b)=\sqrt{x^2-x+1}-(ax+b)\cdot\frac{ \sqrt{x^2-x+1}+(ax+b) }{ \sqrt{x^2-x+1}+(ax+b) }\]

Answer 11

apply L'hospitals now

Answer 12

on your equation @ganeshie8 ?

Answer 13

its getting complicated... one sec

Answer 14

k

Answer 15

u der? @ganeshie8 ?

Answer 16

@pgpilot326 ?

Answer 17

multiply out and gather like terms for the numerator. then split into separate limits? i think this might do it.

Answer 18

\[\lim_{x \rightarrow \infty} \dfrac{ x^2(1-a)+x(-1-2ab)+1-2ab }{ \sqrt{x^2-x+1}+ax+b} \]


the limit is DNE unless \(1-a = 0\)

Answer 19

when a = 1, the limit will be \(\dfrac{1}{2}-b\)

for any other value of a, the limit DNE

Answer 20

how did u get it?

Answer 21

for \(a \ne 1\), numerator's degree is larger than denominator, so as x increases, the expression grows without any bound... consequently the limit wont exist

Answer 22

u applied LH?

Answer 23

\[\sqrt{x^2-x+1}<x,\,\, \forall x>1\]

Answer 24

^^ plugin a=1, then apply LH or divide numerator and denominator by x... anything will do

Answer 25

what did u do? and how can we plug a=1?

Answer 26

\[\lim_{x \rightarrow \infty} \dfrac{ x^2(1-a)+x(-1-2ab)+1-2ab }{ \sqrt{x^2-x+1}+ax+b}\]

we consider two cases :
\(a \ne 1\)
\(a = 1\)

Answer 27

k

Answer 28

for \(a \ne 1\) case, clearly the numerator's degree is 2, and the denominator's degree is 1. so the limit is DNE. (is this convinsing ? )

Answer 29

yes

Answer 30

so the limit for \(a\ne 1\) case is DNE.
lets work \(a = 1\) case

Answer 31

ok nw if a=1 the degree will become same

Answer 32

\(a = 1\) case :

\[\lim_{x \rightarrow \infty} \dfrac{ x(-1-2b)+1-2ab }{ \sqrt{x^2-x+1}+x+b} \]

Answer 33

yes

Answer 34

i got that the limit would exist if a=1 !
but what is the limit?

Answer 35

divide numerator and denominator by x

Answer 36

\[\lim_{x \rightarrow \infty} \dfrac{ (-1-2b)+(1-2ab)/x }{ \sqrt{1-1/x+1/x^2}+1+b/x} \]

Answer 37

take the limit ^^

Answer 38

-1-2b/2

Answer 39

Yep !