University:
Rice University
Course:
MATH 212 | Multivariable Calculus
Academic year:

2023
Views:

405

Pages:

3
Author:

Awayne

Finding Limits Graphically Calculus is split into two rational, differential and integral calculus. The two expressions have an x plus 2 in the denominator. When x is negative 2, the denominator becomes 0, which yields a meaningless result. These two expressions appear quite different from one another. When you divide a number by 0, the result is undefined. The y value does not exist for either of those equations. But what causes that fundamental difference? We will also try to answer this fundamental question: When x approaches negative 2, what does y approach? This will tell us the behavior around negative 2. Why does this function have a vertical asymptote? Why does this one look linear? And then we'll answer that question right now. We can't plug in negative 2 over here because if we plug in negative 2, you actually get 0 over 0, which is 0. Over here, you get 1 over 0, which is undefined. If you take an indeterminant form, the value of your variable can't be determined by the values of its coefficients or its constants. When the function is 0/1, it is undefined. When I refer to the indeterminate form of a function, I mean that the answer to this question cannot be determined. Now, again, the y value of f(-2) is not defined. The statement is true, but I can provide further insight. As y approaches negative 2 from the right, its value gets closer and closer to 1. Let's test this out. As we get closer to negative 2, we're approaching an asymptote. We're not really getting closer to a value. We'll create a table of values for the function x and its derivative f(x). However, we can't plug in negative 2 as an argument because it is not a real number. That's not going to happen. However, we could plug in values around negative 2. For example, –2.1 is far away from 2. To avoid doing a negative 3 or negative 4, we want to go around 2 and then fall back down to 2.1, 2.01, and 2.001. And over here, negative 1.999, negative 1.9, negative 1.99 over here. I have plugged the values into the calculator, and I have obtained the following results. So looking at this, as we are getting closer - so here we're on the left hand side of 2. Over here, we're on the right hand side of 2. We appear to be getting closer and closer, 0.9, 0.99, 0.999. It appears that the y value is approaching 1. And as we approach negative 2 from the other side, the y value varies between 1.1 and 1.001, indicating that it is approaching 1. Thus, as x approaches negative 2, y approaches 1. Here, we answer the question "What is the limit as x approaches negative 2 of x squared plus 5x plus 6 over x plus 2?". And we get 1. When x approaches its limit, what does this approach? At present, we can only do this from a table. Eventually, we will look at algebraic techniques.