Q4. Evaluate limy->1 (y2 - 1) / (y - 1). Q5. Find limx->inf (5x2 + 3x - 1) / (2x3 - 7). Q6. Evaluate limx->0 (1-cosx)/x2 * limx->0 (1-cosx)/x2. Q7. Determine lim_x->-2 (x2+5x+6)/(x+2).

Q4. Evaluate _x 1 (x^3 - 1)/(x - 1) Step 1: Substitute x=1 into the expression. Substituting x=1 gives (1^3 - 1)/(1 - 1) = (0)/(0), which is an indeterminate form. Step 2: Factor the numerator using the difference of cubes formula a^3 - b^3 = (a-b)(a^2+ab+b^2). Here, a=x and b=1, so x^3 - 1 = (x-1)(x^2+x+1). Step 3: Simplify the expression and evaluate the limit. _x 1 ((x-1)(x^2+x+1))/(x-1) = _x 1 (x^2+x+1) Now, substitute x=1: 1^2 + 1 + 1 = 1 + 1 + 1 = 3 The limit is 3. Q5. Find _x (5x^2 + 3x - 1)/(2x^3 - 7) Step 1: Identify the highest power of x in the denominator. The highest power of x in the denominator is x^3. Step 2: Divide every term in the numerator and denominator by x^3. _x (5x^2)/(x^3) + (3x)/(x^3) - (1)/(x^3)(2x^3)/(x^3) - (7)/(x^3) = _x (5)/(x) + (3)/(x^2) - (1)/(x^3)2 - (7)/(x^3) Step 3: Apply the limit property _x (c)/(x^n) = 0 for n > 0. As x , terms like (5)/(x), (3)/(x^2), (1)/(x^3), and (7)/(x^3) all approach 0. (0 + 0 - 0)/(2 - 0) = (0)/(2) = 0 The limit is 0. Q6. Evaluate _x 0 (1 - x)/(x^2) - _x 0 ((1 - x)/(x^2)) Step 1: Recognize the structure of the expression. The expression is in the form A - A, where A = _x 0 (1 - x)/(x^2). Step 2: Evaluate the expression. Since any quantity subtracted from itself is zero, the result is 0. _x 0 (1 - x)/(x^2) - _x 0 ((1 - x)/(x^2)) = 0 The value is 0. Q7. Determine _x -2 (x^2 + 5x + 6)/(x + 2) Step 1: Substitute x=-2 into the expression. Substituting x=-2 gives ((-2)^2 + 5(-2) + 6)/(-2 + 2) = (4 - 10 + 6)/(0) = (0)/(0), which is an indeterminate form. Step 2: Factor the numerator. The quadratic expression x^2 + 5x + 6 can be factored into (x+2)(x+3). Step 3: Simplify the expression and evaluate the limit. _x -2 ((x+2)(x+3))/(x+2) = _x -2 (x+3) Now, substitute x=-2: -2 + 3 = 1 The limit is 1. Send me the next one 📸