One-Sided Limits with Absolute Values and Graph Behavior

The function 3x + 7 is a linear polynomial, which is continuous everywhere. To find the limit as x approaches 0 from the right, we substitute x = 0: 3(0) + 7 = 0 + 7 = 7. Therefore, the limit is 7.

As x approaches 4 from the left, x - 4 approaches 0 from the negative side (a very small negative number). Dividing 1 by a very small negative number yields a very large negative number. Therefore, the limit decreases without bound, which we denote as -∞.

As x approaches 2 from the left, numerator x + 2 → 4 (positive), denominator x - 2 is a small negative number. A positive divided by a small negative gives a large negative value, so the left-hand limit is -∞.

For x approaching -2 from the right (x > -2), x + 2 is positive. The absolute value of a positive number is the number itself, so |x + 2| = x + 2. Therefore, the expression becomes (x + 2)/(x + 2) = 1. So, the limit is 1.

As x approaches 1 from the left, x is slightly less than 1, so 1 - x is a small positive number. The square root of a small positive number is also a small positive number approaching 0. Therefore, the limit is 0.

For x approaching -3 from the left (x < -3), x + 3 is negative. The absolute value of a negative number is its opposite, so |x + 3| = -(x + 3). Therefore, f(x) = [-(x + 3)]/(x + 3) = -1 for x < -3. So, as x approaches -3 from the left, f(x) approaches -1.

The problem states that as x approaches 2 from the left, the y-values approach 3. This means the left-hand limit is 3.

The function 2x - 5 is a linear polynomial, continuous everywhere. We substitute x = -1: 2(-1) - 5 = -2 - 5 = -7. Therefore, the limit as x approaches -1 from the left is -7.

For x > 0 (approaching 0 from the right), |x| = x because x is positive. So, the expression becomes x/x = 1. Therefore, as x approaches 0 from the right, the function value is 1, so the limit is 1.

If the left-hand limit and the right-hand limit at x = c are equal to some number L, then the overall limit as x approaches c exists and equals L. The other options are not necessarily true because the function could still be undefined at c, or not continuous or differentiable even if the limit exists.

Direct substitution gives 0/0. We factor the numerator: x² - 4 = (x - 2)(x + 2). The expression becomes (x - 2)(x + 2)/(x - 2). For x approaching 2 from the right (x > 2), x - 2 is not zero, so we cancel the common factor, leaving x + 2. Now take the limit: 2 + 2 = 4. Therefore, the limit is 4.

If the graph goes downward indefinitely as x approaches 1 from the left, this means the function values decrease without bound. Therefore, the left-hand limit is -∞.

The function 10 - x is a linear polynomial, continuous everywhere. Substitute x = 5: 10 - 5 = 5. Therefore, the limit as x approaches 5 from the right is 5.

The limit as x approaches 0 from the left uses the piece for x < 0, which is f(x) = 3x. Substitute x = 0: 3(0) = 0. Therefore, the left-hand limit is 0.

For a function to be continuous at a point x = c, three conditions must hold: f(c) is defined, the limit of f(x) as x approaches c exists, and the limit equals f(c). The limit exists if and only if the left-hand limit and right-hand limit exist and are equal. Therefore, both one-sided limits must equal the function value at that point for continuity.