Find the value of the following limits: 5. lim (x->2) [1/(x(x-2)2) - 1/(x2-3x+2)]; 6. lim (x->1) [1/(1-x) - 3/(1-x3)]

Here are the solutions to the limit problems. 5. We need to find the value of: _x 2 [ (1)/(x(x-2)^2) - (1)/(x^2-3x+2) ] Step 1: Factor the denominator of the second term. The quadratic x^2-3x+2 factors as (x-1)(x-2). The expression becomes: _x 2 [ (1)/(x(x-2)^2) - (1)/((x-1)(x-2)) ] Step 2: Combine the fractions using a common denominator. The common denominator is x(x-1)(x-2)^2. = _x 2 [ ((x-1))/(x(x-1)(x-2)^2) - (x(x-2))/(x(x-1)(x-2)^2) ] = _x 2 ((x-1) - x(x-2))/(x(x-1)(x-2)^2) Step 3: Simplify the numerator. = _x 2 (x-1 - x^2 + 2x)/(x(x-1)(x-2)^2) = _x 2 (-x^2 + 3x - 1)/(x(x-1)(x-2)^2) Step 4: Evaluate the limit. Substitute x=2 into the numerator: -(2)^2 + 3(2) - 1 = -4 + 6 - 1 = 1. Substitute x=2 into the denominator: 2(2-1)(2-2)^2 = 2(1)(0)^2 = 0. Since the numerator approaches 1 and the denominator approaches 0 from the positive side (because (x-2)^2 is always positive, and x(x-1) is positive near x=2), the limit is (1)/(0^+). The value of the limit is . 6. We need to find the value of: _x 1 ( (1)/(1-x) - (3)/(1-x^3) ) Step 1: Factor the denominator of the second term. The difference of cubes formula is a^3-b^3 = (a-b)(a^2+ab+b^2). So, 1-x^3 = (1-x)(1+x+x^2). The expression becomes: _x 1 ( (1)/(1-x) - (3)/((1-x)(1+x+x^2)) ) Step 2: Combine the fractions using a common denominator. The common denominator is (1-x)(1+x+x^2). = _x 1 ((1+x+x^2) - 3)/((1-x)(1+x+x^2)) Step 3: Simplify the numerator. = _x 1 (x^2 + x - 2)/((1-x)(1+x+x^2)) Step 4: Factor the numerator and simplify. The numerator x^2+x-2 factors as (x+2)(x-1). Also, note that (1-x) = -(x-1). = _x 1 ((x+2)(x-1))/(-(x-1)(1+x+x^2)) Cancel out the common factor (x-1) (since x 1 as x 1): = _x 1 (x+2)/(-(1+x+x^2)) Step 5: Evaluate the limit by substituting x=1. = (1+2)/(-(1+1+1^2)) = (3)/(-(1+1+1)) = (3)/(-3) = -1 The value of the limit is -1. 7. We need to find the value of: _x sqrt(2) (x^4 - 4)/(x^2 + 3sqrt(2)x - 8) Step 1: Check for indeterminate form. Substitute x=sqrt(2) into the numerator: (sqrt(2))^4 - 4 = 4 - 4 = 0. Substitute x=sqrt(2) into the denominator: (sqrt(2))^2 + 3sqrt(2)(sqrt(2)) - 8 = 2 + 3(2) - 8 = 2 + 6 - 8 = 0. This is an indeterminate form (0)/(0), so we need to factor. Step 2: Factor the numerator. The numerator is a difference of squares: x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2). Further factor x^2 - 2 = (x - sqrt(2))(x + sqrt(2)). So, x^4 - 4 = (x - sqrt(2))(x + sqrt(2))(x^2 + 2). Step 3: Factor the denominator. Since substituting x=sqrt(2) into the denominator yields 0, (x-sqrt(2)) must be a factor. We can find the other factor by polynomial division or by inspection: x^2 + 3sqrt(2)x - 8 = (x - sqrt(2))(x + 4sqrt(2)). Step 4: Substitute the factored forms into the limit expression. = _x sqrt(2) (x - sqrt(2))(x + sqrt(2))(x^2 + 2)(x - sqrt(2))(x + 4sqrt(2)) Step 5: Cancel the common factor and evaluate the limit. Cancel out (x - sqrt(2)) (since x sqrt(2) as x sqrt(2)): = _x sqrt(2) (x + sqrt(2))(x^2 + 2)x + 4sqrt(2) Now substitute x=sqrt(2): = (sqrt(2) + sqrt(2))((sqrt(2))^2 + 2)sqrt(2) + 4sqrt(2) = (2sqrt(2))(2 + 2)5sqrt(2) = (2sqrt(2))(4)5sqrt(2) = 8sqrt(2)5sqrt(2) = (8)/(5) The value of the limit is (8)/(5). Last free one today — make it count tomorrow, or type /upgrade for unlimited.