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.
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Here are the solutions to the limit problems. Step 1: Evaluate the first limit. _x 1 (7x^2 - 4x - 3)/(3x^2 - 4x + 1) Substituting x=1 directly gives (7(1)^2 - 4(1) - 3)/(3(1)^2 - 4(1) + 1) = (7-4-3)/(3-4+1) = (0)/(0), which is an indeterminate form. We factor the numerator and the denominator. Since x=1 is a root for both, (x-1) is a factor. Factor the numerator: 7x^2 - 4x - 3 = (x-1)(7x+3). Factor the denominator: 3x^2 - 4x + 1 = (x-1)(3x-1). Now, substitute these factored forms into the limit expression: _x 1 ((x-1)(7x+3))/((x-1)(3x-1)) Cancel out the common factor (x-1) (since x ≠ 1 as x 1): _x 1 (7x+3)/(3x-1) Now, substitute x=1: (7(1)+3)/(3(1)-1) = (7+3)/(3-1) = (10)/(2) = 5 The value of the first limit is 5. Step 2: Evaluate the second limit. _x -2 (x^4 + 5x^3 + 6x^2)/(x^2(x+1) - 4(x+1)) Substituting x=-2 directly gives ((-2)^4 + 5(-2)^3 + 6(-2)^2)/((-2)^2(-2+1) - 4(-2+1)) = (16 - 40 + 24)/(4(-1) - 4(-1)) = (0)/(0), which is an indeterminate form. We need to simplify the expression. Factor the numerator: x^4 + 5x^3 + 6x^2 = x^2(x^2 + 5x + 6) = x^2(x+2)(x+3). Factor the denominator: x^2(x+1) - 4(x+1) = (x+1)(x^2-4) = (x+1)(x-2)(x+2). Now, substitute these factored forms into the limit expression: _x -2 (x^2(x+2)(x+3))/((x+1)(x-2)(x+2)) Cancel out the common factor (x+2) (since x ≠ -2 as x -2): _x -2 (x^2(x+3))/((x+1)(x-2)) Now, substitute x=-2: ((-2)^2(-2+3))/((-2+1)(-2-2)) = (4(1))/((-1)(-4)) = (4)/(4) = 1 The value of the second limit is 1. Step 3: Evaluate the third limit. _x 1 [3]x - 1sqrt(x) - 1 Substituting x=1 directly gives [3]1 - 1sqrt(1) - 1 = (1-1)/(1-1) = (0)/(0), which is an indeterminate form. To simplify, let x = t^6. As x 1, t 1. Then [3]x = [3]t^6 = t^2 and sqrt(x) = sqrt(t^6) = t^3. Substitute these into the limit expression: _t 1 (t^2 - 1)/(t^3 - 1) Factor the numerator using the difference of squares formula a^2-b^2=(a-b)(a+b): t^2 - 1 = (t-1)(t+1). Factor the denominator using the difference of cubes formula a^3-b^3=(a-b)(a^2+ab+b^2): t^3 - 1 = (t-1)(t^2+t+1). Now, substitute these factored forms into the limit expression: _t 1 ((t-1)(t+1))/((t-1)(t^2+t+1)) Cancel out the common factor (t-1) (since t ≠ 1 as t 1): _t 1 (t+1)/(t^2+t+1) Now, substitute t=1: (1+1)/(1^2+1+1) = (2)/(1+1+1) = (2)/(3) The value of the third limit is (2)/(3). Send me the next one 📸