use the formal definition of the limit to prove the statement rigorously. limx →2 x^-2=(1)/(4)

Name: Problem 23, Chapter 2: Calculus for AP
Uploaded: 2020-06-28T20:00:58-08:00
Duration: 10 min 40 s
Channel: Chris Trentman
Description: use the formal definition of the limit to prove the statement rigorously. limx →2 x^-2=(1)/(4)

Use the formal definition of the limit to prove the...

Key Concepts

Epsilon-Delta Definition of a Limit

The epsilon-delta definition of a limit is the cornerstone for rigorous proofs in calculus. It asserts that for every positive number ?, however small, there exists a corresponding positive number ? such that whenever x is within ? of the target value (excluding the target if necessary), the function value f(x) is within ? of the limit L. This precise formulation allows for a strict and unambiguous statement of what it means for a function to approach a limit as x approaches a specific point.

Algebraic Manipulation in Limit Proofs

In epsilon-delta proofs, algebraic manipulation is critical to transform the inequality |f(x) - L| < ? into an inequality that involves |x - a|. Techniques such as factoring, rationalizing, or bounding auxiliary expressions are employed to isolate |x - a| on one side of the inequality, paving the way to explicitly determine an appropriate ? in terms of ?.

Bounding Techniques

Bounding techniques are essential in limit proofs to handle expressions that might otherwise behave unpredictably in the neighborhood of the limit point. By establishing upper bounds for certain components of the function, one can ensure that the chosen ? works uniformly to control |f(x) - L|. This step is crucial in showing that the function's behavior remains contained within the desired ? margin.

Selection of ? in terms of ?

A critical element in any epsilon-delta proof is the careful selection of ?, often expressed as a function of ?, to meet the condition |f(x) - L| < ? whenever |x - a| < ?. This step often involves identifying a relationship between the distance of x from the limit point and the deviation of f(x) from the limit, and then cleverly choosing ? to ensure that this relationship is maintained.

Transcript

00:01 We are asked to prove the following limit using the formal definition of the limit.

00:06 The limit is x approaches 2 on x to the negative second is 1 fourth.

00:15 Let's measure the gap.

00:18 So this is going to be 1 over x squared minus the limit 1 fourth, which is equal to 4 minus x squared over 4 x squared, which is equal to a difference of squares, x minus 2 times x plus 2 over 4 times and then the absolute value of x squared is the same as the absolute value of x squared.

01:06 And we know that x minus 2 is going to lie between 0 and delta.

01:14 So this is going to be less than delta times x plus 2 over 4 times absolute value x squared and how are we going to write x plus 2 and x in terms of delta? well, using our inequality, we have that x minus 2 lies between plus or minus delta, and therefore we have that x is going to lie between 2 plus delta and 2 minus delta.

01:51 To make things simple, let's require that 2 minus delta is greater than or equal to 0.

02:01 And this tells us that delta has to be less than or equal to 2 and then we have that in fact even more restrictive make sure this is over 2 minus delta or over 0 is the delta must be strictly less than 2 and we get that 1 over x is going to be less than 1 over 2 minus delta and greater than 1 over 2 plus delta which is clearly greater than 0.

03:07 So we have that one over the absolute value of x is going to be less than 1 over 2 minus delta and we have that 2 minus delta is going to be greater than 0.

03:35 However, let's pick delta to be instead of just less than 2 less than or equal to 1 and this becomes this is going to be less than or equal to this minus delta is graded than equal to negative 1.

04:06 So 2 minus delta is greater than equal to 1.

04:09 So this is less than equal to 1.

04:30 Or if we didn't want to make this restriction, just that delta is less than 2, we also have that this is going to be strictly less than 1 half.

04:57 That's not true actually.

05:00 We do need to make this less than equal to 1.

05:15 So this negative delta is going to be greater than negative 2.

05:22 So that 2 minus delta is greater than 0.

05:27 In order to invert that, we need to have delta going to be less than equal 1, we'll say.

05:35 We have 2 minus delta is going to be greater than or equal 2.

05:47 2 minus 1, which is 1.

05:50 So we have this going to be less than or equal 2.

06:00 1 over 1, which is just 1.