Limits Involving Trigonometric Function

Example 1: If f(x) = { sin(kx) / x, x < 0 { 3x + 2k^2, x ≥ 0 is continuous at x = 0, find k. Solution: For f(x) to be continuous at x = 0, the left-hand limit and right-hand limit must be equal. lim(x→0-) sin(kx)/x = lim(x→0+) 3x + 2k^2 Using the limit property lim(x→0) sin(x)/x = 1, we get: k = 2k^2 Solving for k: 2k^2 - k = 0 k(2k - 1) = 0 Therefore, k = 0 or k = 1/2. Example 2: If lim(x→0) f(x) = y, then lim(x→0) (4sin(2x) + f(4x)) / (y + f(y)) = ? Solution: x → 0 ⇒ 4x → 0 ⇒ y → 0 Using the given limit and the limit property lim(x→0) sin(x)/x = 1, we get: lim(x→0) (4sin(2x) + f(4x)) / (y + f(y)) = (4*2 + 1) / (0 + 1) = 9 Example 3: If lim(x→0) sin(2ax) / sin(8x) = 14/8, then a = ? Solution: lim(x→0) sin(2ax) / sin(8x) = (2a/8) = 14/8 Solving for a: 2a = 14 a = 7 Squeezing Theorem: If g(x) ≤ f(x) ≤ h(x) for all x, and lim(x→c) g(x) = lim(x→c) h(x) = L, then lim(x→c) f(x) = L. Example: 0 ≤ cos(x^2) ≤ 1 Dividing by x^2: 0 ≤ cos(x^2)/x^2 ≤ 1/x^2 As x approaches 0, 1/x^2 approaches infinity. Therefore, by the squeezing theorem: lim(x→0) cos(x^2)/x^2 = 0 Example 1: lim(x→0) sin(x)/x = 1 Explanation: -1 ≤ sin(x) ≤ 1 Dividing by x: -1/x ≤ sin(x)/x ≤ 1/x As x approaches 0, both -1/x and 1/x approach 0. Therefore, by the squeezing theorem: lim(x→0) sin(x)/x = 0 Example 2: lim(x→0) x^4 * cos(2/π) = 0 Explanation: -1 ≤ cos(2/π) ≤ 1 Multiplying by x^4: x^4 ≤ x^4 * cos(2/π) ≤ x^4 As x approaches 0, both x^4 and x^4 approach 0. Therefore, by the squeezing theorem: lim(x→0) x^4 * cos(2/π) = 0 Example 3: lim(x→∞) e^(-3x) * sin^2(x) = 0 Explanation: 0 ≤ sin^2(x) ≤ 1 Multiplying by e^(-3x): 0 ≤ e^(-3x) * sin^2(x) ≤ e^(-3x) As x approaches infinity, e^(-3x) approaches 0. Therefore, by the squeezing theorem: lim(x→∞) e^(-3x) * sin^2(x) = 0 Example 4: If 3x ≤ f(x) ≤ x^2 + 2 for all x ∈ (-2, 2), find lim(x→0) f(x). Solution: As x approaches 0, both 3x and x^2 + 2 approach 0. Therefore, by the squeezing theorem: lim(x→0) f(x) = 0