How to Prove the Limit of the Floor Function at 1 Using the Epsilon-Delta Method
Understanding the Epsilon-Delta Method
In the study of mathematical analysis, the Epsilon-Delta Limit Definition is a fundamental concept used to rigorously define the limit of a function. In this context, we will demonstrate how to prove that the limit of the floor function ?x? as x approaches 1 does not exist using the epsilon-delta method.
The Epsilon-Delta Proof for ?x? as x Approaches 1
The floor function, denoted as ?x?, represents the largest integer less than or equal to x. For instance, ?1.5? 1 and ?0.9? 0.
To prove that the limit of ?x? as x approaches 1 does not exist, we will show that for any proposed limit L, we can find an ε 0 such that for any δ 0, there exists an x with |x - 1| δ but | ?x? - L | ≥ ε.
Explanation of the Proving Strategy
1. Choosing ε 0.5: First, we choose ε 0.5. This is a strategic choice as it helps us demonstrate a clear contradiction.
2. Examining ?1 - δ? for Various Values of δ: For δ 10-10, 0.1, 0.5, 1.2, 1.9, and 2.1, we observe the values of ?1 - δ? and how they differ from 1 by more than ε 0.5.
3. Interpreting the Results: We notice that for any δ 0, the value ?1 - δ? can be 0, which is different from 1 by more than 0.5. This observation is crucial for proving the non-existence of the limit.
Formal Proof of Non-Existance
By the definition of the limit, we need to find an ε 0 such that for any δ 0, there exists an x within |x - 1| δ such that | ?x? - 1 | ≥ ε.
Step 1: Choose ε 0.5
Since we are using ε 0.5, we need to show that for any δ 0, there exists an x such that | ?x? - 1 | ≥ 0.5.
Step 2: Define x 1 - δ/2
Consider x 1 - δ/2. Notice that 0 δ 1 implies |x - 1| |1 - δ/2 - 1| δ/2 δ. Therefore, x is within |x - 1| δ.
Step 3: Evaluate ?x?
With x 1 - δ/2, we have ?x? ?1 - δ/2? 0 because 1 - δ/2 is less than 1 and greater than 0. Thus, | ?x? - 1 | | 0 - 1 | 1 ≥ 0.5.
Step 4: General Solution for Any δ
We can generalize this by choosing any x from the left of 1, i.e., 0 x 1. For such x, we have ?x? 0, and hence | ?x? - 1 | 1 ≥ 0.5.
Conclusion
The proof shows that no matter how small we choose δ, we can always find an x such that 0 |x - 1| δ and | ?x? - 1 | ≥ 0.5. Therefore, the limit of the floor function at 1 does not exist.
Additionally, notice that any neighborhood of 1 includes points x arbitrarily close to 1 where ?x? 0. This illustrates the discontinuity of the floor function at integer points.