I'm confused about rates of change? Why don't I put this into the limit formula? (Grade 12 Adv Functions)

Question: As you get father from the earth's surface, gravity has less effect on you. For this reason, you actually weigh less at higher altitudes. A person who weighs 55kg can use this function to find out their weight: [6400(55)]/(h+6400), at a specific height (h) above sea evel.

Find the instantaneous rate of change at 12000 ft above sea level.

How my teacher told me to do it:

[w(12000.001-w(1199.999)]/0.002 = 0.00609

Whatever happened to using LIMITS ?

Question

I'm confused about rates of change? Why don't I put this into the limit formula? (Grade 12 Adv Functions)

Question: As you get father from the earth's surface, gravity has less effect on you. For this reason, you actually weigh less at higher altitudes. A person who weighs 55kg can use this function to find out their weight: [6400(55)]/(h+6400), at a specific height (h) above sea evel. 

Find the instantaneous rate of change at 12000 ft above sea level.

How my teacher told me to do it:

[w(12000.001-w(1199.999)]/0.002 = 0.00609


Whatever happened to using LIMITS ?

anonymous · Answer

This Newton's law of universal gravitation i think. 
Well i guess when taking a limit all you need to do is directly plug in the value into h. This would be the limit as h approached 12000 feet?

pottersheep · Answer

Isnt this like a "cheat" way to do limits? What happened to properly using the formula?

anonymous · Answer

He's not giving you the instantaneous rate of change then, hes giving you the average rate of change over a small interval.

If you wanna do a limit you can.

anonymous · Answer

Right i guess so. What he is doing is dw/d(something i dont know)
so dw would be a differential and i dont know what the 0.002 is.
By the way are you in 12th grade AP physics?

anonymous · Answer

the d of something i dont know would not be the 55 kg because that stays constant

pottersheep · Answer

Thanks guys