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Finding δ Given a Specific - Examples 1. Find δ given the following information: f (x) x0 L √ = x−1 = 5 = 1 = 2. We are trying to formally show that lim x→5 √ x − 1 = 2. The value for we are given is telling us how close to our limit, L, we want to be. Our goal is to find a value for δ so that if we stay within the “δ-neighborhood” of x0 , the value of the function will be in the “-neighborhood” of L. In case you have never heard these terms before, here is what they mean: A “δ-neighborhood” around a point x0 is a small interval around x0 . To find it, you simply add δ to x0 (to get the right endpoint), and subtract δ from x0 (to get the left endpoint). Remember that δ is just a very small number, like 0.1. The result is an interval where the left endpoint is x0 − δ, the right endpoint is x0 + δ, and x0 is in the center. Try graphing that on a number line to see if it makes sense. The same thing is true for an “-neighborhood”. Back to our problem. We need to go through the steps we talked about in class to find δ. Step 1: Solve |f (x) − L| < to look like a < x < b. (a and b will just be numbers) √ | x − 1 − 2| √ −1 < x − 1 − 2 √ 1< x−1 1<x−1 2<x < < < < < 1 1 3 9 10. Now we can see that for our problem, a = 2 and b = 10. This tells us what values x has to be in between in order for our function, f (x) to stay within 1 unit (that’s ) of 2 (that’s L). Step 2: Solve |x − x0 | < δ to look like −δ + x0 < x < δ + x0 . Then solve a = −δ + x0 and b = δ + x0 for δ, and choose the smallest! |x − 5| < δ −δ < x − 5 < δ −δ + 5 < x < δ + 5. Next, we need to set up the equations, putting both parts together: −δ + 5 = 2 −δ = −3 δ = 3, and δ + 5 = 10 δ = 5. Choose the smallest of the two. So, δ = 3! We’re done! 2. Find δ given the following information: f (x) x0 L = = = = x2 √ 3 0.1 3. We will use the same steps as in #1 to again find δ. Step 1: |x2 − 3| −0.1 < x2 − 3 2.9 < x2 √ 2.9 < x So a = √ 2.9 and b = √ < < < < 0.1 0.1 3.1 √ 3.1. 3.1. Step 2: √ |x − 3| < δ √ −δ < x − 3 < δ √ √ −δ + 3 < x < δ + 3. Now set up the equations, again putting both parts we’ve done together: √ √ −δ + 3 = 2.9 √ √ −δ = 3 − 2.9 δ = 0.032, and δ+ Therefore, δ = 0.028! Done! √ √ 3 = 3.1 δ = 0.028.