Using the Formal Definition

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Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

limx→9 √(x − 5) = 2

Accepted Answer

Hello. In this video, we are going to prove the limit statement by choosing the correct delta that proves that the given limit exists. So, in order to approach this problem, we need to use the precise definition of a limit. The precise definition of the limit is given as the following. If We have 0, which is less than or equal to X minus A, which is less than or equal to delta. Then we have an inequality F of X minus L, which is less than epsilon. So, what this is saying is that if X minus the value of the limit is less than Delta, then we must have some epsilon inequality, where the function minus the limit is less than epsilon. So, in order to approach this problem, we are going to start by working with the epsilon inequality. Now, in order to set up the epsilon inequality, this is stating that we want to set up the inequality, where we plug in the difference of the function and the value of the limit. Now, in this case here, our function is the square root. Of X minus 4, and the value of the limit we are proving is 3. So the epsilon inequality is going to be the square root of X minus 4. -3, and it's less than Epsilon. Now what we want to do is you want to go ahead and expand this inequality in order to isolate X. By expanding this inequality, we can rewrite the inequality to be negative epsilon less than the square root of X minus 4, -3, which is less than epsilon. What we want to do is we want to isolate X in the middle of this inequality, and to start that, we are going to add 3 to both sides of the epsilon inequality. This will leave us a 3 minus epsilon, less than the square root of X minus 4, which is less than 3 + epsilon. Next, we will go ahead and square both sides of this inequality. That will leave us with 3 minus epsilon square, less than X minus 4, which is less than 3 + epsilon squared. Then what we are going to do is we are going to add 4 to both sides of this equation, or inequality, leaving us with 3 minus epsilon squared plus 4, less than X, less than 3 plus epsilon squared + 4. And now what we want to do is you just want to go ahead and simplify both sides of the inequality. By expanding both sides of the inequality and and combining like terms, we will have 13 minus 6 epsilon. Plus epsilon squared less than X, less than 13 plus 6 epsilon plus epsilon squared. So this is going to be the simplified form of our epsilon inequality. And now what we want to do is you want to repeat this process, but with the delta inequality in order to solve for delta. So, continuing with the delta inequality, recall that Delta is defined as X minus the value of the limit less than Delta. In this case here, the value of the limit that X is approaching is 13. So this is going to be our A. This is going to allow us to set up the inequality, as X minus 13 is less than Delta. And just like the epsilon, we want to expand this inequality and isolate X in the middle of the inequality. We can rewrite the inequality to be negative delta, less than X minus 13, less than delta. And by adding 13 to both sides of this equation, we will get 13 minus delta, less than X, less than 13 plus delta. Now, for whatever value of delta that is going to make this inequality true, we want the values of Epsilon, which are the boundaries of Epsilon, to be within the units of 13 to the left and 13 to the right of delta. So, what we are going to do is we are going to take our values of epsilon and bound them by these values of 13 delta. That is going to give us 13 minus the quantity, 13 minus 6 epsilon plus epsilon squared on the left hand side of the inequality, and on the right hand side, we will have 13 + 6 epsilon plus epsilon squared, and since we want this to be within 13 units, we are going to subtract 13 from this from this quantity. Simplifying the left hand in the left hand quality is going to give us 6 epsilon minus epsilon squared, and on the right hand side, this will simplify to be 6 epsilon plus epsilon squared. So we can simplify the delta inequality to be 6 epsilon, less than epsilon squared, which is less than X, which is less than 6 epsilon plus epsilon squared. These are going to be the values of delta. That make the epsilon inequality true. But since we have two different values of delta, we want delta to be the minimum value between these two boundaries. So, the way that we would write that is that Delta must be the minimum value between 6 epsilon minus epsilon squared and 6 epsilon plus 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.

limx→9 √(x − 5) = 2

Key Concepts

Limit Definition

Square Root Function

Substitution Method

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