lim x-->infinity [ (sqrt x^2 + 4x + 1) - x ] = ?

Question

I'm doing this for practice. The answers in the back of the text says this equals 2. I am a little overwhelmed with this problem and don't know where to begin.

Answer 1

lim [ sqrt(x^2 + 4x + 1) - x ]
x -> infinity

Think of this as a fraction over 1, and multiply top and bottom by the conjugate of the top.

lim [ (sqrt(x^2 + 4x + 1) - x) (sqrt(x^2 + 4x + 1) + x) / (sqrt(x^2 + 4x + 1) + x)
x -> infinity

The top becomes a difference of squares.

lim [ (x^2 + 4x + 1 - x^2) / (sqrt(x^2 + 4x + 1) + x) ]
x -> infinity

Simplify the top.

lim [ (4x + 1) / (sqrt(x^2 + 4x + 1) + x) ]
x -> infinity

Write everything in terms of square roots.

lim [ (sqrt(16x^2) - sqrt(1)) / (sqrt(x^2 + 4x + 1) + sqrt(x^2) ]
x -> infinity

Divide top and bottom by sqrt(x^2).

lim [ (sqrt(16) - sqrt(1/ (16x^2) )) / (sqrt(1 + 4/x + 1/x^2) + 1 ]
x -> infinity

And now, we can solve for the limit directly.

[4 - 0] / [sqrt(1 + 0 + 0) + 1]

4/[sqrt(1) + 1]

4/2

2

Answer 2

lim x-->infinity [ (sqrt x^2 + 4x + 1) - x ]
= lim x-->infinity [4x+1]/[ (sqrt x^2 + 4x + 1) + x ] , multiplied by its conjugate
= lim x-->infinity [4x]/[ (sqrt x^2 ) + x ] , neglected lower degree terms
= 4/2
= 2

Answer 3

fist of all it fairly isn't any longer countless and this shall tend to finite fee as x-> countless the 1st term = sqrt((x-3)^2-8) the above is intuitive and desires a good evidence for comparable to instinct maby bw incorrect enable y = (x^2-6x+a million)^2 y^2 = x^2 - 6x +a million or y^2 - x^2 =a million-6x y-x = (y^2-x^2)/(y+x) = (a million-6x)/(sqrt(x^2-6x+a million) + x) = (a million/x - 6)/(sqrt(a million-6/x+a million/x^2) + a million) as x->countless numerator -> -6 and denominator = 2 so ratio = -3