##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# An Important Limit - Concept

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

An important limit in calculus, which is central to finding derivatives of exponential functions, involves finding the limit as h approaches 0 of a special function. This limit, one of a few **important limits** in Calculus, is the natural logarithm function. The function is useful in Calculus because the derivative of the natural logarithm has a very simple form.

We need a really important result before we can go on and talk about the derivatives of exponential functions and has to do with this limit. So suppose that a is some positive number, some constant I want to find out what the value of this limit is going to be. It's going to be different for every different a value. One way to do this is to make the observation that this limit is going to appear approximately equal to the different quotient itself as long as h is chosen small enough right cause I'm taking the limit as h approaches zero. So let's say for example I choose h to be I don't know 0.1, a to the 0.1 minus 1 over 0.1 this is the sort of thing I'm thinking of. This is a reasonable approximation for this limit and an even better one would involve a smaller value of h like 0.01 here and here. So that's what we're going to do and I want to look specifically at this limit when a is 2 and when a is 3, so what we're going to do right now is compare on the ti84 the limit as h approaches zero of 2 to the h minus 1 over h and ln 2. The limit as h approaches zero 3 to the h minus 1 over h and ln 3. So notice that these numbers are the same, I want to see how these values compare. So let's take a look at that on the TI-84.

Okay we're looking at the TI-84 remember the first thing we wanted to look at was the limit as h approaches zero of 2 to the h minus 1 over h. So I'm going to type in 2 to the h and I have to use alpha carat thatÂ’s h minus 1 and then divided by h, so divided by alpha carat. Okay and thatÂ’s 2 to the h minus 1 over h let's hit enter it equals 1. Now I haven't actually specified what the h value is so it's probably using whatever stored h value it had before I began. So let's actually see what the h value is, it's 1 alright that's pretty good but I actually need h to go to zero so I need something smaller than that. Let's try storing 0.1 so you just type in that number and then the store button, store and then alpha carat thatÂ’s h. So I'm storing 0.1 as h and enter and then instead of typing this whole expression all over again, I'll do second entry, second entry and you just keep hitting second entry and it goes backwards through the entries that you've just executed, now hit this again and now notice I'm getting 0.7177 this is now the new value of the different quotient for h equals 0.1.

Now let do that again let's make smaller value of h so I've typed second entry twice get back to that command. So let me go back and then I'm going to insert second insert throw in a zero. So now it's 0.01 that's being installed as h hit enter and then hit second entry, second entry now let's execute this difference quotient again 0.69555 now remember I'm trying to figure out what limit this is approaching as h gets small. So let's keep letting h get smaller and smaller until we see some kind of limiting behavior, so second entry, hit second entry again. I'm going to now make it 0.000 so second insert another 0 now we have one thousandth and then second entry again, second entry again the different quotient execute 0.693. Okay that's pretty good 0.693 it seems like it's converged to the nearest 100 at this point. So it looks like the value is going to be about 0.693 now we're supposed to compare this value with natural log of 2.

Now this also about 0.693 so this is interesting that these are very very similar to each other. And it turns out that there are one and the same if you let h actually go to zero, this value will be exactly equal to ln2. Now I also wanted to do the same thing for 3, so hit second entry, second entry again I still have remember I have h equals 0.001 all I have to do is go over and change this to 3 to the h. Alright so just type 3 enter so remember h is 0.001 I get 1.099 let's see what ln3 is, ln3 itÂ’s also when you round it to the nearest thousand is also 1.099. So it turns out that this quantity is always going to equal the natural log of this number.

Let's take this back to the board, alright so what did we just learn? What we learned is that these 2 expressions are actually equal, the limit as h approaches zero of 2 to the h minus 1 over h equals ln2. The limit as h approaches zero of 3 to the h minus 1 over h equals ln3 and in general the limit as h approaches zero of a to the h minus 1 over h equals the natural log of this number a.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

#### Related Topics

- Instantaneous Rate of Change 22,815 views
- The Definition of the Derivative 22,503 views
- The Derivative Function 15,838 views
- Derivative Notation 10,861 views
- Derivatives of Linear Functions 16,281 views
- Derivatives of Power Functions 19,336 views
- Derivatives of Polynomial Functions 14,345 views
- Derivatives of Exponential Functions 14,648 views
- Derivatives of Logarithmic Functions 15,359 views
- Average Velocity 26,875 views
- Average Rate of Change 22,054 views
- Instantaneous Velocity 23,523 views

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete