Let $a_n$ be a sequence with $a_n \to \infty$.
We want to consider $e^{\frac{x}{a_n}}$ for $x \in \mathbb{R}$.
By Taylor series expansion, we can write $e^{\frac{x}{a_n}}= 1+\frac{x}{a_n} + o(\frac{x}{a_n})$.
From here, I'm not sure how to interpret $o(\frac{x}{a_n})$. When we write $e^x = 1+x+o(x)$, $o(x)$ means that the remainder $R(x)$ is such that $\lim_{x \to 0} \frac{R(x)}{x} = 0$. But in the expression $o(\frac{x}{a_n})$, we have $x$ that goes to zero and $a_n$ that is indexed by $n$.
So should we write
$o(\frac{x}{a_n}) = o(x) o(\frac{1}{a_n})$, where $o(x)$ is in terms of $x$ going to $0$ and $o(\frac{1}{a_n})$ is in terms of $n$ going to infinity
or write
$o(\frac{x}{a_n}) = o(\frac{1}{a_n})$ since $x$ is fixed?
Thank you,
Why do we have that $x$ goes to zero? $x$ is fixed. It is correct that $o(x/a_n)=o(1/a_n)$ as $n\to\infty$ as constants are irrelevant in the little-o notation.
Taylor series are not needed, anyway. You know $y\mapsto e^y$ is continuous and you know $x/a_n\to0$ as $n\to\infty$, so you know you should get $e^0=1$.