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# One-Sided Limits - Problem 1

###### 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.

You can find some limits by simply plugging in numbers into a function and seeing what number the function seems to be getting closer and closer to. In order to show that the limit exists, you must show that the left-hand and right-hand limits are the same. Start with finding the left-hand limit. To find what f(x) is as it approaches point p from the left, plug in values of x that are less than p. As the values of x get closer and closer to p, look to see what number f(x) gets closer to. To find the right-hand limit, do the same thing, except instead of using values of x that are less than p, use values of x that are greater than p. Then, you will be able to see what f(x) is as it approaches p. So, with this information, you should be able to tell if the limit of f(x) exists as it approaches p, and if it does, you can tell what its limit is.

One of the easiest ways to compute limits, is to calculate them on your calculator. So let’s take a look at the function f(x) equals x² plus 3x minus 10 over x minus 2. I first want to look at the limit as x approaches 2 from the left, of f(x). Let’s take a look at our TI 84. So here’s the TI 84.

First thing I have to do is enter my function. I hit y equals and I want (x² plus 3x minus 10) divided by (x minus 2). So that’s my function. Now I need to go into table set. Actually it’s set up perfectly for me, I want to set it up. I don’t really care about the Table Start or the delta table value, but I want the independent variable set to ask and the dependent set to auto. This is going to make it so that it will allow me to input x values and it will automatically give me the y values.

I hit second table and here I already have a makeshift table. I need my x values to approach 2. You'll see what happens when x equals 2. There’s an error. So 1 is actually a good value. Let me try 1.5. I get 6.5 as my y value. Let me try 1.9. I get 6.9. So I’m trying to figure out what happens as x approaches 2. 1.99, 6.99. So now I’m pretty sure I know what’s happening. It looks like its approaching 7. But I want to try one more. 1.999. Definitely approaching 7. So let’s record this data on our table on the table and we’ll come to our conclusion.

We’re back. I just put my data on the table here. Now remember we wanted to find out what happened as x approaches 2 from the left to f(x). Here I have my values approaching 2 from the left. And it’s pretty clear that the values of f are approaching 7. So I would say that the limit as x approaches 2 from the left of x is 7. Now let’s calculate the limit as x approaches 2 from the right of x. Again we’ll go to the TI 84.

I want to x to approach 2 from the right. So I’m going to start with a value bigger than 2 like 3. I get 8, I get a y value of 8. Let me try 2.1, I get 7.1. How about 2.01, I get 7.01. So it also looks like it’s approaching 7. Let me just try a few more. 2.001, I’m convinced. So let’s record this back on our table and we’ll draw some conclusions. I’ve recorded the data on the table.

Remember I wanted x to approach 2 from the right and that’s what I’m doing here, x is going from 3 to 2.1 to 2.01 and so on, getting closer and closer to 2. But going there from the right, from the more positive values. Look what f(x) is doing, 8, 7.1, 7.01, 7.001. It looks like it’s also approaching 7. So I would say that the limit as x approaches 2 from the right of f(x) is 7.

Now recall when the right hand limit and the left hand limit have the same value, that means the two sided limit exists and has this value. So for part c, what’s the limit as x approaches 2? I can just say 7. Remember when the two one-sided limits exist and they’re equal, the two sided limit has that value, in this case, it's 7.

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###### 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.

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## nadine · 6 months ago

excellent. thanks