# How to evaluate limits of functions involving the little$-o$ notation?

I am studying the conditions for a Counting process to be a Poisson process.

Here, I understand that the "little$$-o$$" notation is defined as follows:

$$\lim_{n\to0}\frac{o(n)}{n}=0$$

Now, I am trying to apply this notation to calculate related limits.

Specifically, what happens when, instead of tending to $$0$$, $$n$$ tends to $$infinity$$, as seen below?

$$\lim_{n\to\infty}$$ $$e^{-n o(1/n)}$$

and

$$\lim_{n\to\infty}$$ $$e^{-n / o(n)}$$

I am new to the "little-$$o$$" notation in terms of how it's used here in the Poisson process. It looks like I am not able to simplify and calculate these limits further without knowing the actual functional form of the functions $$o(1/n)$$ and $$o(n)$$.

Any guidance or insights on how to approach this would be very helpful for me to understand how the function $$o(n)$$ works. Thank you so much!

Edit:

Here's the context behind the above examples:

I have $$n$$ independent Poisson processes: $$[N_k(t): t >= 0]$$, where $$k=1, 2, ..., n$$.

Each of these Poisson processes has a rate of $$1$$.

And $$E$$ is the event that none of the events in any of the processes occur in the time interval $$(0, a_n]$$.

So, $$E$$ $$=$$ $$[N_k(a_n) = 0$$, for all $$k = 1, 2, ..., n]$$

Given this, I am trying to find the below limit:

$$\lim_{n\to\infty}$$ $$P(E)$$

First, I find $$P(E)$$ $$=$$ $$P[N_k(a_n) = 0$$, for all $$k = 1, 2, ..., n]$$

So $$P(E)$$ $$=$$ $$P[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n)) = 0]$$

The sum of $$n$$ independent Poisson processes is also a Poisson process, with rate being equal to the sum of the rates of the $$n$$ processes.

So $$[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n))$$: $$a_n >= 0]$$ is a Poisson process with rate $$(n)$$, because each of the $$n$$ independent Poisson processes has a rate of $$1$$.

This means, $$[N_1(a_n) + N_2(a_n) + ... + N_n(a_n)]$$ is a Poisson random variable with parameter $$(na_n)$$. Is this correct?

Therefore, I get:

$$\lim_{n\to\infty}$$ $$P(E)$$

$$=$$ $$\lim_{n\to\infty}$$ $$P[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n)) = 0]$$

$$=$$ $$\lim_{n\to\infty}$$ $$e^{-na_n}$$

Now, I have $$2$$ cases for which I am trying to evaluate the above limit.

Case $$(i)$$: When $$a_n$$ $$=$$ $$o(1/n)$$

Here, $$\lim_{n\to\infty}$$ $$P(E)$$

$$=$$ $$\lim_{n\to\infty}$$ $$e^{-na_n}$$

$$=$$ $$\lim_{n\to\infty}$$ $$e^{-n o(1/n)}$$

$$=$$ $$e^{-0}$$

$$=$$ $$1$$

Case $$(ii)$$: When $$1/a_n$$ $$=$$ $$o(n)$$

Here, $$\lim_{n\to\infty}$$ $$P(E)$$

$$=$$ $$\lim_{n\to\infty}$$ $$e^{-na_n}$$

$$=$$ $$\lim_{n\to\infty}$$ $$e^{-n / o(n)}$$ --- Because $$1/a_n$$ $$=$$ $$o(n)$$ implies $$a_n = 1/o(n)$$

I am stuck at this point now.

I wonder whether I have been on the right track here. I'd be grateful for any advice or insights on how to proceed further. Thank you so much.

• $f(n)=o(g(n))$ as $n\to\infty$ means $\lim_{n\to\infty}(f(n)/g(n))=0$. Commented Nov 17, 2023 at 8:30
• @GerryMyerson Thank you so much for your reply! This helps me see that $\lim_{n\to\infty}$ $e^{-n o(1/n)}$ $= 1$. For $\lim_{n\to\infty}$ $e^{-n / o(n)}$, I guess the limit depends on how $o(n)$ compares to $n$. If $o(n)$ grows to infinity but at a slower rate than $n$, the denominator in the exponent $−n/o(n)$ will grow, causing the exponent to approach $-infinity$, and the limit of the whole expression to approach $0$. Commented Nov 17, 2023 at 10:29
• I wonder whether we can evaluate this limit $\lim_{n\to\infty}$ $e^{-n / o(n)}$ without more specific information about the growth rate or behavior of $o(n)$ as $n$ approaches infinity. Should I try to work out different cases here, based on the specific behaviour of $o(n)$? Commented Nov 17, 2023 at 10:31
• Have you ever actually seen anyone write $\lim_{n\to\infty}e^{-n/o(n)}$? Commented Nov 17, 2023 at 12:02
• You can think $f(x)=o(g(x))$ $(x\to\text{something})$ by definition means $f(x)=\psi(x)g(x)$ and $\psi(x)\to 0$, $(x\to\text{something})$. So $e^{-n o(1/n)}=e^{-n\psi(n)\frac{1}{n}}=e^{-\psi(n)}$ where $\lim\limits_{n\to\infty}\psi(n)=0$. But $\lim\limits_{n\to\infty}e^{-n o(1/n)}$ is rather weird. $o$ seldom appears in a $\lim$ symbol. Commented Nov 17, 2023 at 12:12

$$o(f(n))$$ is the set of all functions g(n) where the limit of $$g(n) / f(n)$$ is 0. (You need to be careful if f(n) = 0).
Now your examples are very different. $$n \cdot o(1/n)$$ is n, multiplied by a function that becomes smaller and smaller compared to 1/n. So the product becomes smaller and smaller compared to 1.
Now you had $$e^{-n \cdot o(1/n)}$$. That’s e raised to a power that gets smaller and smaller compared to 1, so the limit is $$e^0 = 1$$. In your second case, if you could guarantee that your function that is $$o(n)$$ is always positive, then you would have e raised to larger and larger negative powers with a limit of 0. But sadly, you don’t know that so the values get larger and larger or smaller and smaller.