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September 1, 2014

September 1, 2014

Posted by **Nathan** on Tuesday, October 22, 2013 at 12:43pm.

- Calculus -
**Count Iblis**, Tuesday, October 22, 2013 at 4:51pmsqrt(x^2+6x+3) =

x sqrt(1 + 6/x + 3/x^2)

Using the series expansion:

(1+y)^p = 1 + p y + p (p-1)/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.

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