equivalence of sqrt

[subtle rune]

Ca someone please help me find an equivalence to this function in +oo

[hazy geyser]

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[regal timber]

We have:
(1 + t)^a = 1 + at + a(a - 1)t^2/2 + o(t^2), t -> 0
So, apply this for a = 1/2 by substituting t = x^2 or t = -x^2.

[subtle rune]

I did that but apparently there is a mistake in the quest , the second one is sqrt(x^2 -1)

[regal timber]

Oh. Then that isn't defined around x = 0.

Ah. Or do you need around x = ∞?

[subtle rune]

Yeah I'm working around ♾️

[regal timber]

In that case you can do the following:
√(x^2 + 1) - √(x^2 - 1) = |x|(√(1 + 1/x^2) - √(1 - 1/x^2))
Then you can apply the series above, as 1/x^2 -> 0 as x -> ∞.

[subtle rune]

Ahh thank you +close