A Question About Little-O Notation

Question

How could the following be interpreted: $$ h(t)=\dfrac{1+o(1)}{(2 \pi)^{3/2}\sqrt{t}} \; \; \; \text{as} \; t\downarrow0 \tag{1} $$ Where $o(1)$ is the little-o notation.

From my understanding, if we say that $$f(x)=o(g(x)) \; \; \text{as} \; t \downarrow 0$$ then this means that $$\dfrac{f(x)}{g(x)} \to 0 \; \; \text{as} \; t\downarrow0$$ Now from (1) am I allowed to say that $$ (2 \pi)^{3/2}\sqrt{t}\cdot h(t) -1 =o(1) \; \; \text{as} \; t\downarrow 0 \tag{2}$$ And then by applying the definition of little-o notation to (2), we see that $$ (2 \pi)^{3/2}\sqrt{t}\cdot h(t) -1 \to 0 \; \; \text{as} \; t \downarrow 0 $$

Or should I interpret (1) in a different way?

Answer 1

What you said is correct, and you can also think of it as saying there exists $f$ in $o(1)$ as $t\to0$ such that

$$h(t) = \frac{1+f(t)}{(2\pi)^{3/2}\sqrt{t}}$$

Answer 2

I would write $h(t)=\dfrac{1+o(1)}{(2 \pi)^{3/2}\sqrt{t}} $ as $h(t)=\dfrac{1}{(2 \pi)^{3/2}\sqrt{t}} +o(\dfrac1{\sqrt{t}}) $ so that $h(t)-\dfrac{1}{(2 \pi)^{3/2}\sqrt{t}} =o(\dfrac1{\sqrt{t}}) $ or $\sqrt{t}(h(t)-\dfrac{1}{(2 \pi)^{3/2}\sqrt{t}}) \to 0 $ or $\sqrt{t}h(t)-\dfrac{1}{(2 \pi)^{3/2}} \to 0 $ or $\sqrt{t}h(t)\to\dfrac{1}{(2 \pi)^{3/2}} $.