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I have a rather silly question, I guess. Suppose $(1+x)^{2y}\sim 1$ as $y\to\infty$.

Does this imply $x\sim 1/y$ as $y\to\infty$?

Scuderi
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  • So $x$ is a function of $y$? The answer is no anyways. For example, if $x(y)=0$, this claim is true. Also, $$\lim\limits_{y\to\infty}\left(1+\frac1y\right)^{2y}=e^2\neq0$$ – Rushabh Mehta Feb 01 '21 at 16:23

1 Answers1

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Suppose that $x=x(y) \to 0$ as $y\to +\infty$. Then $$ (1 + x)^{2y} = \exp (2y\log (1 + x)) = \exp (2yx + \mathcal{O}(yx^2 )). $$ So if you want $$ \mathop {\lim }\limits_{y \to + \infty } (1 + x)^{2y} = 1, $$ you need that $$ \mathop {\lim }\limits_{y \to + \infty } xy = \mathop {\lim }\limits_{y \to + \infty } x(y)y = 0, $$ i.e., that $x=o(1/y)$ as $y\to +\infty$. So $x$ must tend to $0$ faster than $1/y$ does.

Gary
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