Question: Yeah, you probably are reading this question and thinking I'm one of those engineers who hasn't learned about the s-plane. I can assure you I'm not! I know that the transfer function of an RC circuit (Vout/Vin) has a REAL pole, with no imaginary component at s= -1/RC + j0. I am curious what happens when we excite an RC circuit at its TRUE pole, which is a decaying exponential with the expression e^(-t/RC).

Of course, the output of an RC Low-pass filter will never blow up to infinity. But what about the ratio of the output to the input, which after all, is the transfer function I originally defined?

Well, let's take this to LTSpice. Below is what I've simulated:
[![enter image description here](https://i.stack.imgur.com/GChDh.png)](https://i.stack.imgur.com/GChDh.png)

I would expect Vout/Vin to be infinity everywhere, however what I see is a ramp. I suspect I'm missing something about how to properly interpret the magnitude response of my transfer function with its true pole as an input. A time-domain view such as what I have should explain something, but I can't understand why it's a ramp.

If anyone has any intuition behind this question, please let me know.

Answer: 
Here is my (very short) answer:

All the calculations in the frequency domain (transfer function, input-and output impedances, pole location,..) do consider STEADY STATE conditions for the circuit under test. Therefore - due to the time dependence of your input - you cannot expect that the circuit will react upon such an input as predicted by the pole analysis.

The observed response of the circuit will be dominated by the time domain behaviour (transient response)...

**Comment** (Edit): In his nice answer KD9PDP has shown that we must switch-over to the time domain - nevertheless, making use of the transfer function - to find the correct result.
It *is* ramping to infinity as you expected. The s-domain calculation is a steady-state calculation, and you have an initial transient (your input exponential was effectively a constant '1' for all t < 0). You have to wait for that to decay (i.e. become negligible) -- e.g. a long (infinite) time.

You can see that after 10 time constants, your 'gain' is already 10.