Piecewise Limits Essentials?

To find lim x→1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→1- f(x). Since for x = 1, f(x) = 2x, as x approaches 1 from the right, we plug in x=1 to get 2*1 = 2. Since 4 is not equal to 2, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = x, as x approaches 0 from the right, we plug in x=0 to get 0. Since both the left-hand limit and the right-hand limit are 0, they are equal. Therefore, the overall limit is 0.

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x ≤ 2, f(x) = 3x - 1, as x approaches 2 from the left, we plug in x=2 to get 3*2 - 1 = 5. Next, the right-hand limit lim x→2+ f(x). Since for x > 2, f(x) = x + 1, as x approaches 2 from the right, we plug in x=2 to get 2 + 1 = 3. Since 5 is not equal to 3, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→-1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-1- f(x). Since for x = -1, f(x) = x², as x approaches -1 from the right, we plug in x=-1 to get (-1)² = 1. Since 2 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→3- f(x). Since for x = 3, f(x) = -x + 5, as x approaches 3 from the right, we plug in x=3 to get -3 + 5 = 2. Since -1 is not equal to 2, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x ≤ 0, f(x) = 4x, as x approaches 0 from the left, we plug in x=0 to get 4*0 = 0. Next, the right-hand limit lim x→0+ f(x). Since for x > 0, f(x) = x³, as x approaches 0 from the right, we plug in x=0 to get 0³ = 0. Since both the left-hand limit and the right-hand limit are 0, they are equal. Therefore, the overall limit is 0.

To find lim x→4 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→4- f(x). Since for x = 4, f(x) = 3, as x approaches 4 from the right, f(x) approaches 3. Since 6 is not equal to 3, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = x, as x approaches 0 from the right, the function approaches 0. Since the left-hand limit is unbounded while the right-hand limit is 0, the behaviors are different. Therefore, the overall limit does not exist.

To find lim x→-2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-2- f(x). Since for x ≤ -2, f(x) = x² + 1, as x approaches -2 from the left, we plug in x=-2 to get (-2)² + 1 = 5. Next, the right-hand limit lim x→-2+ f(x). Since for x > -2, f(x) = 2x + 5, as x approaches -2 from the right, we plug in x=-2 to get 2*(-2) + 5 = 1. Since 5 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→1- f(x). Since for x = 1, f(x) = x², as x approaches 1 from the right, we plug in x=1 to get 1² = 1. Since 3 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x = 2, f(x) = x² - 3, as x approaches 2 from the right, we plug in x=2 to get 2² - 3 = 1. Since 2 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

For the limit of a piecewise function at the point where the definition changes to exist, we calculate the left-hand limit using the left piece and the right-hand limit using the right piece. If these two one-sided limits are equal to the same value, then the overall limit exists and is that value. If they are not equal, the limit does not exist. This condition is necessary because the limit requires the function to approach the same value from both sides.

To determine if the limit at x=a exists, we check if the left-hand limit equals the right-hand limit. Here, both are 5, so they are equal, and the limit exists and is 5. The value of f(a) being 6 does not affect the limit, as the limit is about the behavior as x approaches a, not the value at a itself.

The limit at x=b does not exist if the left-hand limit, calculated using the piece for x approaching b from the left, does not equal the right-hand limit, calculated using the piece for x approaching b from the right. If they are not equal, the function approaches different values from each side. The definition at b or the type of functions do not determine existence; only the agreement of the one-sided limits does.

If a piecewise function uses the same polynomial expression on both sides of the switch point, the function behaves consistently as it approaches the point from either side. Since polynomials are continuous everywhere, the left-hand limit will equal the right-hand limit. Therefore, the overall limit exists.