Finding the limit! Calculus! 10 points!

Can someone help me solve this problem?

lim (x + sqrt(x^2 +2x))
x -> -infinity

Please show all work involved!

rewrite
(x + sqrt(x^2 +2x)) = (x + sqrt(x^2 +2x)) / 1

multiply numerator and denominator by (x - sqrt(x^2 +2x))

(x + sqrt(x^2 +2x)) (x - sqrt(x^2 +2x))/ (x - sqrt(x^2 +2x)) =..............(a+b)(a-b)=a^2-b^2

=( x^2 - (x^2+2x) )/ (x - sqrt(x^2 +2x)) =

= -2x/(x - sqrt(x^2 +2x)) =

collect x^2 inside the sqrt getting

= - 2x /(x - sqrt(x^2(1 + 2/x))) =

remember that sqrt(x^2) is NOT x but |x|
which means that if x --> - infinity it is negative and then |x| = -x

= -2x/(x + x sqrt(1 + 2/x)) =

collect x in the denominator

= - 2x/[x(1 + sqrt(1 + 2/x))] =

simplify x in the num and in the den

= - 2/(1 + sqrt(1 + 2/x)) =............... 2/x --> 0

= - 2/(1 + 1) =

= -1

hope it helps


this is Wolfram's version (more readible but uselessly complicated)
http://img204.imageshack.us/img204/2548/…

We know that x^2 + 2x > 0 for all values of x in the domain (-∞, -2)
That is because x^2 + 2x = x(x + 2) and whenever x < -2, we have x < 0 and (x + 2) < 0 which means that we are multiplying two negative numbers which will always result in a positive number.
When we take the square root of that, it will still approach ∞
However, x -> -∞, which means that we essentially have (-∞) + (∞), which is undefined.
So we basically need to work out what approaches what the quickest.

lim as x -> -∞ of (x + sqrt(x^2 + 2x))
= lim as x -> -∞ of (x + sqrt(x(x + 2)))
As x -> -∞, x(x + 2) -> (x + 1)^2
So we then have:
= lim as x -> -∞ of (x + sqrt((x + 1)^2))
Since we know that x < -1:
= lim as x -> -∞ of (x - (x + 1))
= lim as x -> -∞ of (x - x - 1)
= lim as x -> -∞ of (-1)
= -1

ANSWER = -1