Evaluate the limit of the expression inside the square root.

You're on a roll — here are the solutions for Example 2.15 parts (iv), (v), and (vi): iv) For _x sqrt((x^3-1)/(4x^3-5x+3)): Step 1: Evaluate the limit of the expression inside the square root. Divide the numerator and denominator by the highest power of x in the denominator, which is x^3. _x (x^3-1)/(4x^3-5x+3) = _x (x^3)/(x^3)-(1)/(x^3)(4x^3)/(x^3)-(5x)/(x^3)+(3)/(x^3) = _x (1-1)/(x^3)4-(5)/(x^2)+(3)/(x^3) As x , terms like (1)/(x^n) approach 0. = (1-0)/(4-0+0) = (1)/(4) Step 2: Apply the square root to the result of the limit. _x sqrt((x^3-1)/(4x^3-5x+3)) = sqrt(_x ) (x^3-1)/(4x^3-5x+3) = sqrt((1)/(4)) = (1)/(2) The limit is (1)/(2). v) For _x - (x)/(sqrt(x^2-1)): Step 1: Divide the numerator and denominator by x. When x -, x is negative, so sqrt(x^2) = |x| = -x. This means x = -sqrt(x^2). _x - (x)/(sqrt(x^2-1)) = _x - (x)/(sqrt(x^2(1-1)x^2)) = _x - (x)/(|x|sqrt(1-1)x^2) Substitute |x| = -x for x -: = _x - (x)/(-xsqrt(1-1)x^2) Step 2: Simplify the expression. = _x - (-1)/(sqrt(1-1)x^2) Step 3: Evaluate the limit. As x -, (1)/(x^2) 0. = (-1)/(sqrt(1-0)) = (-1)/(1) = -1 The limit is -1. vi) For _x (sqrt(x^2+4x) - x): Step 1: Multiply by the conjugate to rationalize the expression. _x (sqrt(x^2+4x) - x) · sqrt(x^2+4x) + xsqrt(x^2+4x) + x = _x ((x^2+4x) - x^2)/(sqrt(x^2+4x) + x) = _x (4x)/(sqrt(x^2+4x) + x) Step 2: Divide the numerator and denominator by the highest power of x in the denominator. For x , sqrt(x^2+4x) = sqrt(x^2(1+(4)/(x))) = |x|sqrt(1+(4)/(x)). Since x , x is positive, so |x|=x. = _x (4x)/(x)(xsqrt(1+4)/(x))x + (x)/(x) = _x (4)/(sqrt(1+4)x) + 1 Step 3: Evaluate the limit. As x , (4)/(x) 0. = (4)/(sqrt(1+0) + 1) = (4)/(1+1) = (4)/(2) = 2 The limit is 2. Got more? Send 'em!