Homework Help

Limit x->infinity (sqrroot(x+1)+sqrroot(x-1))

user profile pic

rockarnab | Student, Grade 11 | eNoter

Posted November 6, 2011 at 5:13 PM via web

dislike 0 like

Limit x->infinity (sqrroot(x+1)+sqrroot(x-1))

2 Answers | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 6, 2011 at 6:11 PM (Answer #1)

dislike 0 like

Multiply squareroot(x+1)+squareroot(x-1) by the conjugate squareroot(x+1)-squareroot(x-1) and you'll get a difference of squares.

[squareroot(x+1)+squareroot(x-1)][squareroot(x+1)-squareroot(x-1)]=[squareroot(x+1)]^2-[squareroot(x-1)]^2

[squareroot(x+1)+squareroot(x-1)][squareroot(x+1)-squareroot(x-1)]=x+1-x+1

[squareroot(x+1)+squareroot(x-1)][squareroot(x+1)+squareroot(x-1)]=2

 limit [squareroot(x+1)+squareroot(x-1)]=limit [2/(squareroot(x+1)-squareroot(x-1))] = 2/infinite = 0

Answer: The limit of the addition is 0 if x approaches to infinite.

user profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 6, 2011 at 8:47 PM (Answer #2)

dislike 0 like

`lim_(x->oo) sqrt(x+1) + sqrt(x-1) `

`==> lim_(x->oo) (sqrt(x+1)+sqrt(x-1))X (sqrt(x+1)-sqrt(x-1))/(sqrt(x+1)-sqrt(x-1))`

`==> lim_(x->oo) ((x+1) - sqrt(x^2-1) +sqrt(x^2-1) - (x-1))/(sqrt(x+1)-sqrt(x-1))`

`==> lim_(x->oo) (x+1-x+1)/(sqrt(x+1)-sqrt(x-1))`

`==> lim_(x->oo) 2/(sqrt(x+1)-sqrt(x-1)) = 2/oo = 0`

`==>lim_(x->oo) sqrt(x+1) + sqrt(x-1) = 0`

``

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes