What is the limit of the square root of x^2 + 4x + 1 subtracted by x, as x approaches positive infinity?

Question

Hello ! I've tried to solve for this problem, but I can't seem to find an answer wherein the limit exists. I would appreciate assistance with regards to determining what I did wrong for my solutions. Thank you in advance !

Answer 1

If the problem asked for the limit of the square root of x^2+4x+4 - x, what would you think the answer was?

I would guess that you would say x+2 - x = 2, so it is always 2, and the limit is 2.

We shall use the standard trick to generate this result.

We multiply the numerator by conjugate/conjugate.

We then get (√(x²+4x+1) - x)*(√(x²+4x+1) + x)/(√(x²+4x+1) + x) =

(4x+1)/(√(x²+4x+1) + x)

For x > 0,

(4x+1)/(√(x²+4x+4) + x) < (4x+1)/(√(x²+4x+1) + x) < (4x+1)/(√(x²+2x+1) + x) , or

(4x+1)/(2x+2) < (4x+1)/(√(x²+4x+1) + x) < (4x+1)/(2x+1)

2 - 3/(2x+2) < (4x+1)/(√(x²+4x+1) + x) < 2 - 1/(2x+1)

lim x -> ∞ 2 - 3/(2x+2) = 2

lim x -> ∞ 2 - 1/(2x+1) = 2

Thus, lim x -> ∞ (4x+1)/(√(x²+4x+1) + x) = 2

lim x -> ∞ √(x²+4x+1) - x = 2

Answer 2

I am assuming you mean "divided by x" so the expression is

Lim _x-->∞ { ( √( x² + 4x + 1) ) / x }

Bring the x into the square root and divide each term. Since the x is going under the radical sign, we divide each term by x² :

= Lim _x-->∞ { √( 1 + 4/x + 1/x²) }

As x-->∞:

1 stays constant

4/x goes to zero

1/x² goes to zero

Therefore, Lim x-->∞ { ( √( x2 + 4x + 1) ) / x } = 1