# find the limits of f(x) when x is tend to infinity Watch

1. as x tends to be infinity, what's the limits of (x^2+x)^(1/2)-x
2. laurent series expansion of f(x) is f(x) = 2^(-1) + [8x]^(-1) + [16x^2]^(-1) + ...

As x -> infinity, x^-1, x^-2 etc. approach 0, so f(x) approaches 2^-1 = 1/2
3. Bash in numbers of x that are very high and very low and then you'll see where they tend to.
4. (Original post by FailedMyMocks)
Bash in numbers of x that are very high and very low and then you'll see where they tend to.
You'll need to (most likely) justify why the limiting value is the one that you've said.

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Updated: April 24, 2016
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