One-Sided Limits Using Exponential, Trig, and Special Limit Techniques

The left-hand limit is determined by the values of f(x) for x

For x near 2 but not equal, (x - 2)² is a small positive number. Its reciprocal 1/(x - 2)² is a large positive number regardless of approaching from left or right. Approaching from the left yields values that grow without bound, so the left-hand limit is +∞.

Use the standard limit lim as t→0 sin(t)/t = 1. Rewrite sin(2x)/x = [sin(2x)/(2x)] * 2. As x→0 from the left, sin(2x)/(2x) → 1, so the whole expression tends to 2. The one-sided approach does not change the value, so the left-hand limit equals 2.

For x ≠ 1, rationalize numerator: (√(x) - 1)/(x - 1) = (√(x) - 1)/[(√(x) - 1)(√(x) + 1)] = 1/(√(x) + 1). As x→1^- , √(x) → 1, so the expression tends to 1/(1 + 1) = 1/2. Therefore the left-hand limit equals 1/2.

Use the small-angle approximation or known limit: (1 - cos(x))/x² → 1/2 as x→0. More formally, use the Taylor expansion cos(x) ≈ 1 - x²/2 + ... so 1 - cos(x) ≈ x²/2, dividing by x² gives 1/2. The right-hand limit equals 1/2.

For x > 3, x - 3 is positive, so |x - 3| = x - 3. The ratio equals 1 for x > 3. Approaching 3 from the right the expression remains 1, so the right-hand limit is 1.

For x approaching 0 from the left, 1/x is a large negative number. Exponentiating gives e^{large negative} which tends to 0. Therefore the left-hand limit is 0.

Factor numerator: x² - 4 = (x - 2)(x + 2). Then the expression becomes (x - 2)(x + 2)/(x - 2)² = (x + 2)/(x - 2) for x ≠ 2. For x approaching 2 from the left, x + 2 → 4 (positive) while x - 2 is a small negative number, so the quotient (x + 2)/(x - 2) is a large negative number. So the limit is -∞.

The value of a one-sided limit tells you what the function approaches, not necessarily what the function equals at that point. The point (3, 5) could be a hole, or the function could be defined as a different value at x = 3 (a jump discontinuity).

For small x, use series expansions to compare growth up to order x. Expand 1/sin(x) using sin(x) ≈ x - x³/6 + ... so 1/sin(x) ≈ 1/x * 1/(1 - x²/6 + ...) ≈ 1/x * (1 + x²/6 + ...). Therefore 1/x - 1/sin(x) ≈ 1/x - [1/x * (1 + x²/6)] = 1/x - 1/x - x/6 + ... = -x/6 + higher-order terms. As x→0⁺ this tends to 0, not -1/6. The constant term is 0; the limit is 0. Therefore the correct answer is A. 0.

For x approaching -1 from the right, x > -1 so x + 1 is positive. Then |x + 1| = x + 1 and the ratio (x + 1)/(x + 1) = 1 for x ≠ -1. As x→-1⁺ the expression remains 1, so the right-hand limit is 1.

Tangent near zero behaves like its argument: tan(x) ≈ x. As x approaches 0 from the left, tan(x) approaches 0 from the left as well, but the limit value is 0. Therefore the left-hand limit equals 0.

A limit describes the behavior of a function as it gets closer to a specific x-value, not the value of the function at that exact point. A function can have a hole at x = a (making it undefined there) while the graph still approaches a specific height from the right, meaning the limit exists.

Factor numerator: x³ - 1 = (x - 1)(x² + x + 1). For x ≠ 1 the expression simplifies to x² + x + 1. Taking the right-hand limit as x→1⁺ gives 1 + 1 + 1 = 3. So the one-sided limit equals 3.

Use the known expansion or derivative at zero: the limit of (e^x - 1)/x as x→0 equals the derivative of e^x at 0, which is 1. Approaching from the right gives the same value, so the right-hand limit is 1.