Finding the Limit of f(x) = 2x + 1 as x → 2 using ε-δ
Step 1: Identify the Function and Point of Interest
We are dealing with the function f(x) = 2x + 1 and we want to find its limit as x approaches 2.
Step 2: Calculate the Limit
First, let’s find lim(x → 2) (2x + 1). Plugging in x = 2 into the function, we get 2 × 2 + 1 = 5.
Step 3: Set Up ε-δ Conditions
We want to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |f(x) - 5| < ε.
Step 4: Express |f(x) – 5|
We have |f(x) – 5| = |2x + 1 – 5| = |2x – 4| = 2|x – 2|.
Step 5: Find δ
We want 2|x – 2| < ε, which simplifies to |x - 2| < ε/2.
So, we can choose δ = ε/2.
Step 6: Verify the Conditions
If 0 < |x - 2| < δ, then |f(x) - 5| = 2|x - 2| < 2δ = ε.
Final Result
By the ε-δ definition, lim(x → 2) (2x + 1) = 5.
Numerical Illustration of the Limit of f(x) = 2x + 1 as x → 2 using ε-δ
Step 1: Choose an ε Value
Let’s choose ε = 0.1 as an example.
Step 2: Calculate δ
From our previous work, we know that δ = ε/2. So, δ = 0.1/2 = 0.05.
Step 3: Pick x Values Close to 2
We’ll pick x values that are within 0.05 units of 2, say x = 1.95 and x = 2.05.
Step 4: Evaluate f(x)
For x = 1.95, f(x) = 2 × 1.95 + 1 = 3.9 + 1 = 4.9
For x = 2.05, f(x) = 2 × 2.05 + 1 = 4.1 + 1 = 5.1
Step 5: Check the ε Condition
We want to see if |f(x) – 5| < ε.
For x = 1.95, |4.9 – 5| = 0.1 which is equal to ε.
For x = 2.05, |5.1 – 5| = 0.1 which is also equal to ε.
Step 6: Conclusion
Our chosen x values satisfy the ε-δ condition, illustrating that the limit of f(x) as x → 2 is indeed 5.