Function composition and little-o notation

Question

Let's define $o(g(x))$ as usually:

$$ \forall x \ne a.g(x) \ne 0 \\ f(x) = o(g(x)) \space \text{when} \space x \to a \iff \lim_{x \to a} \frac{f(x)}{g(x)}=0 $$

By Taylor's expansion at 0, we have

$\tag{1} e^u = 1 + u + o(u) \text{ when } u \to 0 \iff \lim_{u \to 0} \frac{e^u - 1 - u}{u} = 0$

How does the little-o look like when we do Taylor expansion near 0, but of function $e^{x \log x}$?

If I just replace $u$ with $x \log x$ in $(1)$, then I get

$\tag{2} e^{x \log x} = 1 + x \log x + o(x \log x) \iff \lim_{x \log x \to 0} \frac{e^{x \log x} - 1 - x \log x}{x \log x} = 0$

One way we could proceed is by knowing that

$\tag{3} \lim_{x \to 0+} x \log x = 0$

Then let $f(u) = \frac{e^u - 1 - u}{u}$ and $g(x) = x \log x$

Then by $(1)$ we know that $\lim_{u \to 0} f(u) = 0$ and by $(3)$ we know that $\lim_{x \to 0+} g(x) = 0$, but it's not completely obvious to me that:

$\lim_{x \to 0+} f(g(x)) = 0$

Is it possible to prove that $f(u)$ above satisfies the conditions needed for the final composition? Is there a simpler way, without introducing new definitions, to look at the problem, such that it is clear that the $o(x \log x)$ is taken at the point $x \to 0+$?

Thanks!

Answer 1

If $\lim_{x \to a} g(x) = b$ and $\lim_{x \to b} f(x) = c$, then $$\lim_{x \to a} f(g(x)) = c$$ whenever either one of two conditions is satisfied. Either

• $f(b) = c$ (i.e., $f$ is continuous at $b$), or
• there is a punctured neighborhood of $a$ on which $g$ is not equal to $b$.

To claim $\lim_{x \to a} f(g(x)) = c$, we need to show that for every neighborhood $W$ of $c$ there is some neighborhood $U$ of $a$ such that for all $x \in U$ with $x \ne a$, we have $f(g(x)) \in W$.

Let $W$ be a neighborhood of $c$. Because $\lim_{x \to b} f(x) = c$, there is some neighborhood $V$ of $b$ such that for all $t \in V$ with $t \ne b$, we have $f(t) \in W$. Because $\lim_{x \to a} g(x) = b$, there is some neighborhood $U$ of $a$ such that if $x \in U$ and $x \ne a$, then $g(x) \in V$. If $g(x) \ne b$, then it satisfies the earlier condition on $V$, so $f(g(x)) \in W$ as desired. But there is a problem when $g(x) = b$. The limit for $f$ does not tell us anything about the behavior of $f(b)$. It could be $c$, it could be some other value, or it could be undefined. The two conditions listed solve this problem in different ways:

• If $f(b) = c$, then it does not matter whether $g(x) = b$. For those $x \in U$ with $g(x) \ne b$, we have $f(g(x)) \in W$ by the two given limits. If $g(x) = b$, we have $f(g(x)) = f(b) = c$. Since $c \in W$, in all cases we have $f(g(x)) \in c $, which shows $\lim_{x \to a} f(g(x)) = c$.
• If there is some punctured neighborhood $N$ of $a$ for which $g(x)\ne b$ for all $x \in N$, then we can choose $U' = N \cap U$ as our punctured neighborhood of $a$ for the final limit. If $x \in U'$, then $x \in N$, so $g(x) \ne b$, and $x \in U$, so $g(x) \in V$, and together, these mean $f(g(x)) \in W$, so $\lim_{x \to a} f(g(x)) = c$.

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In your case $g(x) = x\log x$, and we know that $\lim_{x \to 0+} g(x) = 0$, but $g(x) \ne 0$ for $x \in (0,1)$. So $g$ satisfies the second condition. Thus we know that $\lim_{x \to 0+}f(g(x)) =0$.