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Suppose that I am given a (very complicated) function $f(x)$ which I know is infinitely differentiable (and, therefore, continuous) over all real $x$. Taylor expansion around $x=0$ yields the following very simple form:

$$f(x)=ax+\mathcal{O}(x^2)$$ with constant $a>0$.

Now suppose I have another function:

$$g(y)=\operatorname{Exp}[y^{b}(c/y-f(1/y))]$$ where $b$ and $c$ are constants.

I am interested in proving the following statement:

If $b>1$ and $c <a$, then $\lim_{y\rightarrow\infty}g(y)=0$.

This might be a very easy question, as the statement seems to be trivialy true, however, how do I justify using the limit in the context of $f(x)$ being expressed as a Taylor series? If this is not true, then what conditions would I need to prove for $f(x)$ such that it is true?

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up vote 1 down vote accepted

It's true indeed. Since $f(x) = ax + O(x^2)$ as $x \to 0$ you have

$$ \begin{align*} y^b\Bigl(c/y - f(1/y)\Bigr) &= y^b\Bigl(c/y - a/y + O(1/y^2)\Bigr) \\ &= (c-a)y^{b-1}\Bigl(1 + O(1/y)\Bigr) \longrightarrow -\infty \end{align*} $$

as $y \to \infty$, so that

$$ \exp\left\{y^b\Bigl(c/y - f(1/y)\Bigr)\right\} \longrightarrow 0 $$

as $y \to \infty$.

To justify the first limit you could note that $1 + O(1/y) > 1/2$ for $y$ large enough, so that $$(c-a)y^{b-1}\Bigl(1 + O(1/y)\Bigr) < \frac{c-a}{2} \,y^{b-1}.$$

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Do you mean the "greater than" sign (">") in your last line? – M.B.M. Feb 25 '13 at 2:49
@M.B.M. No; $(c-a)y^{b-1}$ is negative, so if $A > 1/2$ then $A(c-a)y^{b-1} < (1/2)(c-a)y^{b-1}$. – Antonio Vargas Feb 25 '13 at 2:52
Ahh, yeah, nevermind! :) – M.B.M. Feb 25 '13 at 3:02

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