Evaluate the limit _x 3 (x2 - 9)/(x2 - 2x - 3) and determine if the function f(x) is continuous at x=2. If not, identify the type of discontinuity.

Here are the solutions to the questions on limits and continuity: Step 1: Evaluate the limit _x 3 (x^2 - 9)/(x^2 - 2x - 3) Step 1: Substitute x=3 into the expression to check for an indeterminate form. (3^2 - 9)/(3^2 - 2(3) - 3) = (9 - 9)/(9 - 6 - 3) = (0)/(0) Since we have an indeterminate form, we need to simplify the expression. Step 2: Factor the numerator and the denominator. The numerator is a difference of squares: x^2 - 9 = (x-3)(x+3). The denominator is a quadratic: x^2 - 2x - 3 = (x-3)(x+1). Step 3: Substitute the factored forms into the limit expression and cancel common factors. _x 3 ((x-3)(x+3))/((x-3)(x+1)) Since x 3, x ≠ 3, so we can cancel the (x-3) terms. _x 3 (x+3)/(x+1) Step 4: Substitute x=3 into the simplified expression. (3+3)/(3+1) = (6)/(4) = (3)/(2) The limit is (3)/(2). Step 2: Determine if the following function is continuous at x=2. If not, identify the type of discontinuity. f(x) = (x^2 - 4)/(x-2) & if x ≠ 2 \\ 3 & if x = 2 For a function to be continuous at x=c, three conditions must be met: 1. f(c) must be defined. 2. _x c f(x) must exist. 3. _x c f(x) = f(c). Step 1: Check if f(2) is defined. From the definition of the function, f(2) = 3. So, f(2) is defined. Step 2: Evaluate the limit _x 2 f(x). For x ≠ 2, f(x) = (x^2 - 4)/(x-2). _x 2 (x^2 - 4)/(x-2) Factor the numerator: x^2 - 4 = (x-2)(x+2). _x 2 ((x-2)(x+2))/(x-2) Cancel the (x-2) terms since x ≠ 2. _x 2 (x+2) Substitute x=2: 2+2 = 4 So, _x 2 f(x) = 4. The limit exists. Step 3: Compare f(2) and _x 2 f(x). We have f(2) = 3 and _x 2 f(x) = 4. Since f(2) ≠ _x 2 f(x), the function is not continuous at x=2. Step 4: Identify the type of discontinuity. Since the limit _x 2 f(x) exists but is not equal to f(2), this is a removable discontinuity. If we redefined f(2) to be 4 instead of 3, the discontinuity would be removed. The function is not continuous at x=2. It has a removable discontinuity. Step 3: Find the value of k that makes the function continuous at x=1. g(x) = 2x + k & if x 1 \\ 5 - x & if x > 1 For g(x) to be continuous at x=1, the left-hand limit, the right-hand limit, and the function value at x=1 must all be equal. _x 1^- g(x) = _x 1^+ g(x) = g(1) Step 1: Find g(1). Using the first part of the piecewise function for x 1: g(1) = 2(1) + k = 2 + k Step 2: Find the left-hand limit, _x 1^- g(x). For x < 1, g(x) = 2x + k. _x 1^- (2x + k) = 2(1) + k = 2 + k Step 3: Find the right-hand limit, _x 1^+ g(x). For x > 1, g(x) = 5 - x. _x 1^+ (5 - x) = 5 - 1 = 4 Step 4: Set the limits and function value equal to solve for k. For continuity, we must have: 2 + k = 4 Subtract 2 from both sides: k = 4 - 2 k = 2 The value of k that makes the function continuous at x=1 is 2. Last free one today — make it count tomorrow, or type /upgrade for unlimited.