as x tends to be infinity, what's the limits of (x^2+x)^(1/2)-x
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find the limits of f(x) when x is tend to infinity Watch
- Thread Starter
- 16-04-2016 20:05
- 17-04-2016 02:37
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
- 24-04-2016 13:42
Bash in numbers of x that are very high and very low and then you'll see where they tend to.
- 24-04-2016 13:44