Posted by Nathan on .
Evaluate the limit. as x approaches infinity, lim sqrt(x^2+6x+3)x

Calculus 
Count Iblis,
sqrt(x^2+6x+3) =
x sqrt(1 + 6/x + 3/x^2)
Using the series expansion:
(1+y)^p = 1 + p y + p (p1)/2 y^2 + ...
for p = 1/2 gives:
sqrt(1 + 6/x + 3/x^2) = 1 + 3/x +
O(1/x^2)
Where the O(1/x^2) means that there exists an R and a constant c such that for x larger than R, the difference between the square root and (1+3/x) becomes less than c/x^2.
We then have:
x sqrt(1 + 6/x + 3/x^2)  x =
3 + O(1/x)
This means that the limit for x to infinity is 3.