Find the Values Of the following Limits

Here are the values of the given limits: 1) _x 2 (3-x)/(2+x) Step 1: Substitute x=2 into the expression. _x 2 (3-x)/(2+x) = (3-2)/(2+2) Step 2: Simplify the expression. (1)/(4) The value of the limit is (1)/(4). 2) _x 3 (x^2-9)/(x-3) Step 1: Factor the numerator using the difference of squares formula, a^2-b^2=(a-b)(a+b). _x 3 ((x-3)(x+3))/(x-3) Step 2: Cancel out the common factor (x-3). _x 3 (x+3) Step 3: Substitute x=3 into the simplified expression. 3+3 = 6 The value of the limit is 6. 3) _x 2 (x^3-8)/(x^2-4) Step 1: Factor the numerator using the difference of cubes formula, a^3-b^3=(a-b)(a^2+ab+b^2), and the denominator using the difference of squares formula. _x 2 ((x-2)(x^2+2x+4))/((x-2)(x+2)) Step 2: Cancel out the common factor (x-2). _x 2 (x^2+2x+4)/(x+2) Step 3: Substitute x=2 into the simplified expression. (2^2+2(2)+4)/(2+2) = (4+4+4)/(4) = (12)/(4) = 3 The value of the limit is 3. 4) _x 0 sqrt(1+x^2)-sqrt(1-x)x^2 Step 1: Multiply the numerator and denominator by the conjugate of the numerator, which is sqrt(1+x^2)+sqrt(1-x). _x 0 sqrt(1+x^2)-sqrt(1-x)x^2 · sqrt(1+x^2)+sqrt(1-x)sqrt(1+x^2)+sqrt(1-x) Step 2: Expand the numerator using (a-b)(a+b)=a^2-b^2. _x 0 ((1+x^2)-(1-x))/(x^2(sqrt(1+x^2)+1-x)) Step 3: Simplify the numerator. _x 0 (1+x^2-1+x)/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x^2+x)/(x^2(sqrt(1+x^2)+1-x)) Step 4: Factor out x from the numerator. _x 0 (x(x+1))/(x^2(sqrt(1+x^2)+1-x)) Step 5: Cancel out a common factor of x. _x 0 (x+1)/(x(sqrt(1+x^2)+1-x)) Upon re-evaluating, there was a mistake in the previous step. The x in the denominator is x^2, so after canceling one x, there is still an x left in the denominator. Let's re-do from Step 3. Corrected Step 3: Simplify the numerator. _x 0 (1+x^2-1+x)/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x^2+x)/(x^2(sqrt(1+x^2)+1-x)) Corrected Step 4: Factor out x from the numerator. _x 0 (x(x+1))/(x^2(sqrt(1+x^2)+1-x)) Corrected Step 5: Cancel out a common factor of x. _x 0 (x+1)/(x(sqrt(1+x^2)+1-x)) This still results in (1)/(0) if we substitute x=0. Let's check the original problem again. The problem is _x 0 sqrt(1+x^2)-sqrt(1-x)x^2. Numerator: (1+x^2) - (1-x) = x^2+x. So, (x^2+x)/(x^2(sqrt(1+x^2)+1-x)) = (x(x+1))/(x^2(sqrt(1+x^2)+1-x)) = (x+1)/(x(sqrt(1+x^2)+1-x)). This limit does not exist because the denominator approaches 0 while the numerator approaches 1. Let's re-examine the problem statement or my interpretation. The problem is _x 0 sqrt(1+x^2)-sqrt(1-x)x^2. If the problem was _x 0 sqrt(1+x^2)-sqrt(1-x)x, then after multiplying by conjugate, we would get (x^2+x)/(x(sqrt(1+x^2)+1-x)) = (x(x+1))/(x(sqrt(1+x^2)+1-x)) = (x+1)/(sqrt(1+x^2)+1-x). Substituting x=0 would give (1)/(sqrt(1)+1) = (1)/(2). However, the denominator is x^2. Let's re-check the algebra carefully. _x 0 sqrt(1+x^2)-sqrt(1-x)x^2 Multiply by conjugate: _x 0 sqrt(1+x^2)-sqrt(1-x)x^2 · sqrt(1+x^2)+sqrt(1-x)sqrt(1+x^2)+sqrt(1-x) = _x 0 ((1+x^2)-(1-x))/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (1+x^2-1+x)/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x^2+x)/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x(x+1))/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x+1)/(x(sqrt(1+x^2)+1-x)) As x 0, the numerator approaches 0+1=1. As x 0, the denominator approaches 0 · (sqrt(1+0^2)+sqrt(1-0)) = 0 · (1+1) = 0. Since the numerator approaches a non-zero number and the denominator approaches zero, the limit does not exist. More specifically, if x 0^+, the denominator is x · (positive number) 0^+, so the limit is +. If x 0^-, the denominator is x · (positive number) 0^-, so the limit is -. Since the left and right limits are different, the limit does not exist. Let's consider if L'Hopital's Rule is applicable. The form is (0)/(0). Let f(x) = sqrt(1+x^2)-sqrt(1-x) and g(x) = x^2. f'(x) = (1)/(2sqrt(1+x^2))(2x) - (1)/(2sqrt(1-x))(-1) = (x)/(sqrt(1+x^2)) + (1)/(2sqrt(1-x)) g'(x) = 2x _x 0 (f'(x))/(g'(x)) = _x 0 (x)/(sqrt(1+x^2)) + (1)/(2sqrt(1-x))2x = _x 0 ( (x)/(2xsqrt(1+x^2)) + (1)/(4xsqrt(1-x)) ) = _x 0 ( (1)/(2sqrt(1+x^2)) + (1)/(4xsqrt(1-x)) ) As x 0, the first term approaches (1)/(2sqrt(1)) = (1)/(2). The second term approaches (1)/(0), which means it does not exist. So, L'Hopital's Rule also indicates the limit does not exist. It's possible the problem intended a different denominator or a different numerator. Given the image, x^2 is clearly the denominator. I will state that the limit does not exist. Step 1: Multiply the numerator and denominator by the conjugate of the numerator. _x 0 sqrt(1+x^2)-sqrt(1-x)x^2 · sqrt(1+x^2)+sqrt(1-x)sqrt(1+x^2)+sqrt(1-x) Step 2: Simplify the numerator using (a-b)(a+b)=a^2-b^2. _x 0 ((1+x^2)-(1-x))/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x^2+x)/(x^2(sqrt(1+x^2)+1-x)) Step 3: Factor out x from the numerator and cancel one x with the denominator. _x 0 (x(x+1))/(x^2(sqrt(1+x^2)+1-x)) = _x 0 (x+1)/(x(sqrt(1+x^2)+1-x)) Step 4: Evaluate the limit. As x 0, the numerator approaches 0+1=1. The denominator approaches 0 · (sqrt(1+0^2)+sqrt(1-0)) = 0 · (1+1) = 0. Since the numerator approaches a non-zero value and the denominator approaches zero, the limit does not exist. The value of the limit is Does Not Exist. 5) _x 1 (1-x^2)/(1+x) Step 1: Factor the numerator using the difference of squares formula, a^2-b^2=(a-b)(a+b). _x 1 ((1-x)(1+x))/(1+x) Step 2: Cancel out the common factor (1+x). _x 1 (1-x) Step 3: Substitute x=1 into the simplified expression. 1-1 = 0 The value of the limit is 0. Send me the next one 📸