Let f be defined by f(x)= 4-x2 & if x 1 2+x2 & if x > 1. Find each of the following limits if they exist: (i) x 1^- f(x) (ii) x 1^+ f(x) (iii) _x 1 f(x)

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Asked on June 6, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-06-06T15:59:59.908Z

Here are the points of discontinuity for each function: a) For the function f(x) = (x^2+x-6)/(x+3) & if x ≠ -3 \\ -5 & if x = -3 : Step 1: Check the function value at x=-3. f(-3) = -5 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. For x ≠ -3, we can simplify the expression: (x^2+x-6)/(x+3) = ((x+3)(x-2))/(x+3) = x-2 Now, find the limit: _x -3 f(x) = _x -3 (x-2) = -3-2 = -5 The limit exists. Step 3: Compare the function value and the limit. Since _x -3 f(x) = -5 and f(-3) = -5, we have _x -3 f(x) = f(-3). For all other values of x, the function is a polynomial (after simplification), which is continuous. Therefore, the function f(x) is continuous everywhere. There are no values of x for which f(x) is not continuous. b) For the function h(x) = (x^2+x-6)/(x+3) & if x ≠ -3 \\ 1 & if x = -3 : Step 1: Check the function value at x=-3. h(-3) = 1 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. As shown in part (a), for x ≠ -3, (x^2+x-6)/(x+3) = x-2. _x -3 h(x) = _x -3 (x-2) = -3-2 = -5 The limit exists. Step 3: Compare the function value and the limit. Since _x -3 h(x) = -5 and h(-3) = 1, we have _x -3 h(x) ≠ h(-3). Therefore, the function h(x) is not continuous at x=-3 because the limit of the function as x approaches -3 does not equal the function's value at x=-3. c) For the function g(x) = (1)/(x+3) & if x ≠ -3 \\ 3 & if x = -3 : Step 1: Check the function value at x=-3. g(-3) = 3 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. Consider the one-sided limits: _x -3^+ (1)/(x+3) = + _x -3^- (1)/(x+3) = - Since the left-hand limit and the right-hand limit are not equal (and both are infinite), the limit _x -3 g(x) does not exist. Therefore, the function g(x) is not continuous at x=-3 because the limit of the function as x approaches -3 does not exist. d) For the function f(x) = (|x|)/(x): Step 1: Check the function value at x=0. The expression (|x|)/(x) involves division by x. If x=0, the denominator is zero. f(0) = (|0|)/(0) = (0)/(0) The function f(x) is undefined at x=0. Therefore, the function f(x) is not continuous at x=0 because the function is undefined at this point. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-06-06T15:59:59.908Z

Here are the points of discontinuity for each function: a) For the function f(x) = (x^2+x-6)/(x+3) & if x ≠ -3 \\ -5 & if x = -3 : Step 1: Check the function value at x=-3. f(-3) = -5 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. For x ≠ -3, we can simplify the expression: (x^2+x-6)/(x+3) = ((x+3)(x-2))/(x+3) = x-2 Now, find the limit: _x -3 f(x) = _x -3 (x-2) = -3-2 = -5 The limit exists. Step 3: Compare the function value and the limit. Since _x -3 f(x) = -5 and f(-3) = -5, we have _x -3 f(x) = f(-3). For all other values of x, the function is a polynomial (after simplification), which is continuous. Therefore, the function f(x) is continuous everywhere. There are no values of x for which f(x) is not continuous. b) For the function h(x) = (x^2+x-6)/(x+3) & if x ≠ -3 \\ 1 & if x = -3 : Step 1: Check the function value at x=-3. h(-3) = 1 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. As shown in part (a), for x ≠ -3, (x^2+x-6)/(x+3) = x-2. _x -3 h(x) = _x -3 (x-2) = -3-2 = -5 The limit exists. Step 3: Compare the function value and the limit. Since _x -3 h(x) = -5 and h(-3) = 1, we have _x -3 h(x) ≠ h(-3). Therefore, the function h(x) is not continuous at x=-3 because the limit of the function as x approaches -3 does not equal the function's value at x=-3. c) For the function g(x) = (1)/(x+3) & if x ≠ -3 \\ 3 & if x = -3 : Step 1: Check the function value at x=-3. g(-3) = 3 The function is defined at x=-3. Step 2: Evaluate the limit as x -3. Consider the one-sided limits: _x -3^+ (1)/(x+3) = + _x -3^- (1)/(x+3) = - Since the left-hand limit and the right-hand limit are not equal (and both are infinite), the limit _x -3 g(x) does not exist. Therefore, the function g(x) is not continuous at x=-3 because the limit of the function as x approaches -3 does not exist. d) For the function f(x) = (|x|)/(x): Step 1: Check the function value at x=0. The expression (|x|)/(x) involves division by x. If x=0, the denominator is zero. f(0) = (|0|)/(0) = (0)/(0) The function f(x) is undefined at x=0. Therefore, the function f(x) is not continuous at x=0 because the function is undefined at this point. That's 2 down. 3 left today — send the next one.

Let f be defined by f(x)= 4-x2 & if x 1 2+x2 & if x > 1. Find each of the following limits if they exist: (i) x 1^- f(x) (ii) x 1^+ f(x) (iii) _x 1 f(x)

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Let f be defined by f(x)= 4-x2 & if x 1 2+x2 & if x > 1. Find each of the following limits if they exist: (i) x 1^- f(x) (ii) x 1^+ f(x) (iii) _x 1 f(x)

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