Using the Formal Definition

...

Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

lim x→0 √(4 − x) = 2

Accepted Answer

Hello. In this video, we are going to be proving the limit statement by choosing a value of delta that correctly proves the limit. So, in order to approach this problem, we are going to need to use the precise definition of the limit. The precise definition is given as the following. If we have an inequality, X minus A less than or equal to delta, then there must be an epsilon inequality defined as F of X minus L less than epsilon. So, how do we start this problem and how do we utilize this definition? Well, in order to start this problem, we are going to use our epsilon inequality as our starting point, since we are trying to solve for delta in terms of Epsilon. What the epsilon inequality states is that we are going to take our function and subtract the value of the limit from it. Now, in order to set up this inequality, we will take our function, the square root of 5 minus X, and subtract the value of 2 from it. That will give us the absolute value of the square root 5 minus X minus 2 less than epsilon. Now what we want to do is we want to expand and simplify this inequality to get our epsilon bounds. So, the first thing we are going to do is we are going to rewrite this inequality. Without the absolute value symbol, and that is going to give us negative epsilon, less than the square root of 5 minus X, minus 2, which is less than epsilon. Now what we want to do is we want to isolate this X in the middle of the inequality. We will go ahead and start by adding 2 to both sides, and that will give us 2 minus epsilon, less than the square root of 5 minus X, which is less than 2 plus epsilon. We will square both sides of this inequality, leaving us with 2 minus epsilon squared, less than 5 minus X, which is less than 2 + epsilon squared. Then what we are going to do is we are going to subtract 5 to both sides of the inequality. This will leave us with 2 minus epsilon squared, minus 5, less than negative X, which is less than 2 + epsilon squared. -5. Now, before we divide by -1, let's go ahead and expand both sides of this inequality. By expanding both sides of this inequality, we will have -1 minus 4 epsilon plus epsilon squared is less than -X, which is less than -1 + 4 epsilon plus epsilon squared. Now, if we divide by -1, keep in mind that that is going to change the signs of every term in our inequality, but it is also going to flip the direction of our inequality. So if we divide the entire inequality by -1, we are going to end up with 1 minus 4 epsilon minus epsilon squared, which is less than X, which is less than 1 + 4 multiplied by epsilon minus epsilon squared. So what we've created are the boundaries of our Epsilon neighborhood for the limit. We're going to set this to the side for now, and now what we want to do is we want to determine the boundaries of our delta neighborhood. In order to do this, we are going to expand the delta inequality from the precise definition of a limit. Now, in order to create the inequality, we want to pick the value of A, which is the value that X is approaching in the limit. In this limit, X is approaching the value of 1, so A is going to be 1. That is going to give us X minus 1 is less than Delta, and just like the Epsilon inequality, we are going to expand this inequality without the absolute value symbol. We can rewrite the inequality to be negative delta is less than X minus 1, which is less than delta, and by adding 1 to both sides of this equation or inequality, we get 1 minus delta is less than X, which is less than 1 + delta. Now what this is saying is that for whatever delta value we pick, Delta is going to be within 1 unit to the left and to the right of X, and that means that we want our epsilon boundaries to be one unit to the left and to the right of this delta value. So, we are going to replace the deltas. On both sides of this inequality with our epsilon boundaries. For the left hand side, we will have 1 minus the left epsilon boundary, which is going to be 1 minus 4 epsilon minus epsilon squared. And on the right hand side, since we want our epsilon boundary to be within 1 unit of X, that means we are going to take our right epsilon boundary, 1 + 4 epsilon minus epsilon squared, and bound it within 1 unit, so we are going to subtract one from it. Now what we are going to do is we are going to simplify these two sides of the inequality. On the left hand side, by expanding and combining like terms, this will simplify to be 4 epsilon plus epsilon squared. This will be less than X, and on the right hand side, by expanding and combining like terms, we have 4 epsilon minus epsilon squared. So, what this is, is that these are going to be the values of delta that cause the limit to be true. But since we have two values of delta, we want delta to be the minimum value between these two values. So, the way that we will state the value of delta is that delta must be the minimum value between 4 epsilon plus epsilon squared, and 4 epsilon minus epsilon squared. And this is going to be the solution to the problem. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

lim x→0 √(4 − x) = 2

Key Concepts

Limit Definition

Square Root Function

Limit Properties

Watch next