I have two similar problems that I need help completing. Please show all your work.

Question: Find the limit L. Then use the å-ä definition to prove the limit is L.

1. lim (2x+5)
x->3
2. lim 3
x->6
Thank you for your anticipated help!!

**Note**

Please use the Epsilon-Delta Definition.

1)

f(x) = 2 x + 5

The limit for x to 3 is 11, assuming the function is continuous. To prove this, consider the open interval

I = (11-epsilon, 11+epsilon)

for some arbitrary epsilon
We want to find a delta > 0 such that the function f maps the open interval

J =(3-delta, 3+delta)

into I. That would then prove that for any arbitry epsilon>0 such a delta indeed exists. This means that no matter how small you take epsilon, there is always an interval containing x = 3 that will be mapped by f into the interval of size 2 epsilon containing 11.

If we solve:

2 x + 5 = y

for x, we get:

x = (y-5)/2

If we take y = 11 +/- epsilon, we get:

x = 3 +/- epsilon/2

So, we can take delta = epsilon/2

To find the limit L for each problem, we substitute the given value of x into the function expression and simplify. Let's first consider the first problem:

1. lim (2x + 5) as x approaches 3.

To find the limit, substitute x = 3 into the expression:
lim (2(3) + 5) = lim (6 + 5) = lim 11 = 11.

Therefore, the limit L for the first problem is 11.

Now, let's move on to using the ε-δ definition to prove the limit is L.

The ε-δ definition states that: For any ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε.

We need to show that for any given ε > 0, we can find a δ > 0 that satisfies the definition. Let's proceed with the proof for each problem:

1. lim (2x + 5) as x approaches 3.

Let's assume ε > 0 is given. Now, we need to find a δ > 0 such that if 0 < |x - 3| < δ, then |(2x + 5) - 11| < ε.

To find such a δ, let's manipulate the inequality:
|(2x + 5) - 11| < ε
|2x - 6| < ε
|2(x - 3)| < ε
2|x - 3| < ε

To satisfy this inequality, we can set δ = ε/2.

Now, if 0 < |x - 3| < δ, then |2(x - 3)| < 2(ε/2) = ε.

Thus, we have proven that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |(2x + 5) - 11| < ε, which satisfies the ε-δ definition.

Therefore, the limit of (2x + 5) as x approaches 3 is 11.

Now let's move on to the second problem:

2. lim 3 as x approaches 6.

Using the same approach as before, we substitute x = 6 into the expression:
lim 3 = 3.

Therefore, the limit L for the second problem is 3.

To prove the limit using the ε-δ definition, we assume ε > 0 and need to find a δ > 0 such that if 0 < |x - 6| < δ, then |3 - 3| < ε.

Since |3 - 3| = 0, and for any ε > 0, 0 is always less than ε, we can choose any δ > 0, meaning the choice of δ does not depend on ε.

Thus, we have proven that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 6| < δ, then |3 - 3| < ε, which satisfies the ε-δ definition.

Therefore, the limit of 3 as x approaches 6 is 3.

I hope this explanation helps! Let me know if you have any more questions.