If your program logic allows you to do, you can pulse it or pwm it at or below its operating voltage.
Even with a very short duty-cycle, you're still over-driving the PIC output pin, and the LED
during peak-pulse times. I for sure don't recommend you do this without the series resistor
when directly driving the LED with a PIC output pin.

With a transistor driving the LED, you may be able to get away with high peak-pulsed
current levels, for short durations, but I would never try this directly driving the LED from
the PIC output pin.

I assume a 5v LED would have a built in resistor as standard LEDs are around 2v.
I wouldn't assume the LED had a built-in resistor. I would check the data sheet for the LED
first. If you have an LED with a forward voltage drop of 2V, and a MAX forward current of
10mA, then just subtract 2V from 5V for the working voltage, then /10mA to find the series
resistor value.

I.E. 5V - 2V = 3V / 10mA = 300-ohm series resistor you'll need to operate the LED at 5V
with 10mA current through the LED.

What if you need to operate this same LED with a 12V supply? Same thing applies. 12V -2V
= 10V. 10V / 10mA = 1K. So you just pop in a 1K resistor. Most LEDs don't care what the
voltage is, you just have to limit current through the LED at whatever voltage you're using.

Now, if you want to use PWM to dim or brighten the LED, your peak-pulse current levels will
not exceed the LED max DC current levels during the peak-pulse periods.

If you drive the LED with a 50-50 PWM duty cycle, your time average current will be
50% of 10mA, but your peak-pulse-current will still be the full 10mA.

If you exceed the MAX drive levels for the PIC or LED, even for brief periods, you can
pretty much count on failure at some point.

Seems pretty silly when you could have prevented total failure, or degredation over a
period of time with a 1-cent resistor...;o}