Evaluate the following limits: 9. lim x->0 x2 cos(1/x) 10. lim x->-2 |x+2|/(x+2)

Here are the solutions to the limit problems. 9. We need to evaluate the limit: _x 0 x^2 ((1)/(x)) Step 1: Understand the behavior of ((1)/(x)). For any x 0, the value of the cosine function is always between -1 and 1. -1 ((1)/(x)) 1 Step 2: Apply the Squeeze Theorem. Multiply all parts of the inequality by x^2. Since x^2 0 for all real x, the direction of the inequalities remains the same. -x^2 x^2 ((1)/(x)) x^2 Step 3: Evaluate the limits of the bounding functions as x 0. _x 0 (-x^2) = -(0)^2 = 0 _x 0 (x^2) = (0)^2 = 0 Step 4: Conclude using the Squeeze Theorem. Since both the lower bound ( -x^2 ) and the upper bound ( x^2 ) approach 0 as x 0, the limit of the function in between must also be 0. The limit is 0. 10. We need to evaluate the limit: _x -2 (|x+2|)/(x+2) Step 1: Define the absolute value function for x near -2. The absolute value |x+2| behaves differently depending on whether x+2 is positive or negative. |x+2| = x+2 & if x+2 0 x -2 \\ -(x+2) & if x+2 < 0 x < -2 Step 2: Calculate the left-hand limit. As x -2^- (meaning x approaches -2 from values less than -2), x+2 is negative. So, |x+2| = -(x+2). _x -2^- (-(x+2))/(x+2) = _x -2^- (-1) = -1 Step 3: Calculate the right-hand limit. As x -2^+ (meaning x approaches -2 from values greater than -2), x+2 is positive. So, |x+2| = x+2. _x -2^+ (x+2)/(x+2) = _x -2^+ (1) = 1 Step 4: Compare the left-hand and right-hand limits. Since the left-hand limit (-1) is not equal to the right-hand limit (1), the limit does not exist. The limit is Does Not Exist. Last free one today — make it count tomorrow, or type /upgrade for unlimited.