How do you prove that the limit of #sqrt(x+1) = 2 # as x approaches 3 using the epsilon delta proof?

we have:

Now consider that:

and clearly:

so:

we have:

To prove that the limit of sqrt(x+1) as x approaches 3 is equal to 2 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 3| < delta, then |sqrt(x+1) - 2| < epsilon.

Let's begin the proof:

Given epsilon > 0, we need to find a suitable delta > 0.

We start by manipulating the expression |sqrt(x+1) - 2| < epsilon:

|sqrt(x+1) - 2| < epsilon |sqrt(x+1) - 2| * |sqrt(x+1) + 2| < epsilon * |sqrt(x+1) + 2| |x+1 - 4| < epsilon * |sqrt(x+1) + 2| |x - 3| < epsilon * |sqrt(x+1) + 2|

Now, we can see that if we choose delta = epsilon * |sqrt(x+1) + 2|, then:

|x - 3| < delta |x - 3| < epsilon * |sqrt(x+1) + 2|

Since we want to prove the limit as x approaches 3, we can assume that 0 < |x - 3| < delta. Therefore, we can substitute delta into the inequality:

|x - 3| < epsilon * |sqrt(x+1) + 2| epsilon * |sqrt(x+1) + 2| < epsilon * |sqrt(x+1) + 2|

Hence, we have shown that for any given epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon * |sqrt(x+1) + 2|) such that if 0 < |x - 3| < delta, then |sqrt(x+1) - 2| < epsilon. This satisfies the definition of the limit, proving that the limit of sqrt(x+1) as x approaches 3 is indeed equal to 2.