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

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`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`

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`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`

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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.

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