speed is $$\frac{\text{distance}}{\text{time interval}}$$

but at the event horizon of a black hole, time interval becomes $0$.

Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, $0.001$ flashes/s $0.00001$ flashes/s, and decreasing as it approaches at the event horizon.

The number of flashes per second is a time interval observed by us (people far away from the black hole).

So, flashlight's one second is our 100 seconds.

Flashlight at the event horizon sees the light traveling at a speed of light (in vacuum) for 1 seconds (can be 10 seconds or $x$ seconds), we'll see a light traveling at a speed of light for 0 seconds.

From the formula:

$$ \text{speed} \times \text{time interval} = \text{distance} $$

:)

Although light has constant speed of $299,792,458$ [m/s] in a local inertial frame neglecting regarding elements (i.e., in vacuum), distance that light travels at the event horizon becomes 0 (it is traveling in a constant (and finite) speed though).

Light never decelerates, or changes direction. It travels straight "out" with respect to the person falling in, but straight "in" with respect to us, the observers outside the black hole (insofar as much as the relevant coordinate systems say) radially with a constant speed, yet we the external observers never observe it penetrating the event horizon, as if it had stopped there.

Isn't it weird?

speed is $$\frac{\text{distance}}{\text{time interval}}$$

but at the event horizon of a black hole, time interval becomes $0$.

Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, $0.001$ flashes/s $0.00001$ flashes/s, and decreasing as it approaches at the event horizon.

The number of flashes per second is a time interval observed by us (people far away from the black hole).

So, our one second is flashlight's 0.00001 seconds.

Flashlight at the event horizon sees the light traveling at a speed of light (in vacuum) for 1 seconds (can be 10 seconds or $x$ seconds), we'll see a light travelling at a speed of light for 0 seconds.

From the formula:

$$ \text{speed} \times \text{time interval} = \text{distance} $$

:)

Although light has constant speed of $299,792,458$ [m/s] in a local inertial frame neglecting regarding elements (i.e., in vacuum), distance that light travels at the event horizon becomes 0 (it is travelling in a constant (and finite) speed though).

Light never decelerates, or changes direction. It travels straight "out" with respect to the person falling in, but straight "in" with respect to us, the observers outside the black hole (insofar as much as the relevant coordinate systems say) radially with a constant speed, yet we the external observers never observe it penetrating the event horizon, as if it had stopped there.

Isn't it weird?

speed is $$\frac{\text{distance}}{\text{time interval}}$$

but at the event horizon of a black hole, time interval becomes $0$.

Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, $0.001$ flashes/s $0.00001$ flashes/s, and decreasing as it approaches at the event horizon.

The number of flashes per second is a time interval observed by us (people far away from the black hole).

So, flashlight's one second is our 0.001 second.
Flashlight at the event horizon sees the light traveling at a speed of light (in vacuum) for 1 seconds (can be 10 seconds or $x$ seconds), we'll see a light traveling at a speed of light for 0 seconds.

from the formula

$$ \text{speed} \times \text{time interval} = \text{distance} $$

:)

Although light has constant speed of $300,000,000$ [m/s], distance that light travels at the event horizon becomes 0 (it is traveling in a constant (and finite) speed though).

Light never decelerates, or changes direction. It travels straight out radially with a constant speed, yet amount of time it travels is 0 for us, the observers!

It is weird.

speed is $$\frac{\text{distance}}{\text{time interval}}$$

but at the event horizon of a black hole, time interval becomes $0$.

Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, $0.001$ flashes/s $0.00001$ flashes/s, and decreasing as it approaches at the event horizon.

The number of flashes per second is a time interval observed by us (people far away from the black hole).

So, flashlight's one second is our 100 seconds.

Flashlight at the event horizon sees the light traveling at a speed of light (in vacuum) for 1 seconds (can be 10 seconds or $x$ seconds), we'll see a light traveling at a speed of light for 0 seconds.

From the formula:

$$ \text{speed} \times \text{time interval} = \text{distance} $$

:)

Although light has constant speed of $299,792,458$ [m/s] in a local inertial frame neglecting regarding elements (i.e., in vacuum), distance that light travels at the event horizon becomes 0 (it is traveling in a constant (and finite) speed though).

Light never decelerates, or changes direction. It travels straight "out" with respect to the person falling in, but straight "in" with respect to us, the observers outside the black hole (insofar as much as the relevant coordinate systems say) radially with a constant speed, yet we the external observers never observe it penetrating the event horizon, as if it had stopped there.

Isn't it weird?