Algebraic Limits: Core Concepts and Techniques

The function f(x) = 2x² - 3x + 1 is a polynomial. According to the Direct Substitution Property, we can evaluate the limit by simply plugging in x=5. We compute f(5) = 2*(5)² - 3*(5) + 1 = 2*25 - 15 + 1 = 50 - 15 + 1 = 36. Thus, the limit is 36.

Direct substitution gives (1-1)/(1-1)=0/0. We factor both the numerator and the denominator. The numerator is a difference of cubes: x³ - 1 = (x - 1)(x² + x + 1). The denominator is a difference of squares: x² - 1 = (x - 1)(x + 1). The limit becomes lim_(x→1) [ (x - 1)(x² + x + 1) ] / [ (x - 1)(x + 1) ]. Cancel the common factor (x - 1), provided x ≠ 1. We get lim_(x→1) (x² + x + 1) / (x + 1). Now substitute x=1: (1² + 1 + 1) / (1 + 1) = (1+1+1)/2 = 3/2.

We start by noticing that direct substitution gives (3² - 9)/(3 - 3) = 0/0, which is indeterminate. The expression can be factored. The numerator factors as (x - 3)(x + 3). So the original limit becomes lim_(x→3) [(x - 3)(x + 3)] / (x - 3). We cancel the common factor of (x - 3), provided x ≠ 3. This simplifies the expression to lim_(x→3) (x + 3). Now we evaluate by direct substitution: 3 + 3 = 6. Therefore, the limit is 6.

We check the one-sided limits. As x approaches 4 from the right (x→4+), the numerator approaches 2 and the denominator approaches 0 through positive values. Thus, the quotient grows without bound towards positive infinity (+∞). As x approaches 4 from the left (x→4⁻), the numerator approaches 2 and the denominator approaches 0 through negative values. Thus, the quotient decreases without bound towards negative infinity (-∞). Since the left-hand limit and right-hand limit are not equal, the two-sided limit does not exist.

Direct substitution gives ( (0+2)² - 4 ) / 0 = (4-4)/0 = 0/0, an indeterminate form. We simplify the expression algebraically. First, expand the numerator: (x+2)² - 4 = (x² + 4x + 4) - 4 = x² + 4x. So the limit becomes lim_(x→0) (x² + 4x) / x. We factor the numerator: x(x + 4). The expression becomes lim_(x→0) [x(x + 4)] / x. Canceling the common factor of x (x ≠ 0), we get lim_(x→0) (x + 4). Evaluating by direct substitution gives 0 + 4 = 4.

Direct substitution gives (4-4)/(16-16)=0/0. We have a radical expression. To simplify, we multiply the numerator and denominator by the conjugate of the numerator: (4+√x). The limit becomes limx→16 [ (√(x) - 4)(√(x) + 4) ] / [ (x - 16)(√(x) + 4) ]. The numerator simplifies using the difference of squares: (√(x))² - (4)² = x - 16. So we have limx→16 (x - 16) / [ (x - 16)(√(x) + 4) ]. Cancel the common factor (x - 16), provided x ≠ 16. We get limx→16 1 / (√(x) + 4). Now substitute x=16: 1 / (√(16) + 4) = 1 / (4 + 4) = 1/8.

We use the limit laws. The limit of a difference is the difference of the limits, and the limit of a constant times a function is the constant times the limit. So, lim_(x→2) [ 2f(x) - g(x) ] = 2 * lim_(x→2) f(x) - lim_(x→2) g(x). Substitute the given values: 2*(5) - (-3) = 10 + 3 = 13.

Direct substitution gives ((-1)² + 3*(-1) + 2) / ((-1)³ + 1) = (1 -3 +2) / (-1+1) = 0/0. We factor both polynomials. The numerator: x² + 3x + 2 = (x+1)(x+2). The denominator is a sum of cubes: x³ + 1 = (x+1)(x² - x + 1). The limit becomes lim_(x→-1) [ (x+1)(x+2) ] / [ (x+1)(x² - x + 1) ]. Cancel the common factor (x+1), provided x ≠ -1. We get lim_(x→-1) (x+2) / (x² - x + 1). Substitute x = -1: (-1+2) / ((-1)² - (-1) + 1) = (1) / (1 +1 +1) = 1/3.

This is a complex fraction. Combine the terms in the numerator into a single fraction. The common denominator for 1/(x+2) and 1/2 is 2(x+2). So, 1/(x+2) - 1/2 = [2 - (x+2)] / [2(x+2)] = [2 - x - 2] / [2(x+2)] = (-x) / [2(x+2)]. The original limit becomes lim_(x→0) [ (-x) / (2(x+2)) ] / x = lim_(x→0) (-x) / [2(x+2) * x] = lim_(x→0) (-x) / [2x(x+2)]. Cancel the common factor of x (x ≠ 0), to get lim_(x→0) -1 / [2(x+2)]. Now substitute x=0: -1 / [2*(0+2)] = -1/(2*2) = -1/4.

To evaluate limits at infinity for rational functions, we divide both the numerator and the denominator by the highest power of x in the denominator, which is x². This gives lim_(x→∞) (3 - 2/x + 5/x²) / (4 + 1/x + 1/x²). As x→∞, terms with x in the denominator (like 2/x or 5/x²) approach 0. Thus, the limit is (3 - 0 + 0) / (4 + 0 + 0) = 3/4. (Note: Since the degrees of the numerator and denominator are equal, this confirms the shortcut that the limit is simply the ratio of the leading coefficients, 3/4).

This involves an absolute value, so we consider one-sided limits. As x approaches 3 from the right (x→3^+), x > 3, so |x-3| = (x-3). The expression becomes (x-3)/(x-3) = 1. Thus, the right-hand limit is 1. As x approaches 3 from the left (x→3⁻), x

We use the limit law for quotients, provided the limit of the denominator is not zero. Since lim_(x→c) g(x) = 2 ≠ 0, we have lim_(x→c) [ f(x) / g(x) ] = [lim_(x→c) f(x)] / [lim_(x→c) g(x)] = 8 / 2 = 4.

Direct substitution gives ((-2)³ + 8) / ((-2)² - 4) = (-8+8)/(4-4)=0/0. Factor both. Numerator: sum of cubes: x³ + 8 = (x+2)(x² - 2x + 4). Denominator: difference of squares: x² - 4 = (x+2)(x-2). The limit becomes lim_(x→-2) [ (x+2)(x² - 2x + 4) ] / [ (x+2)(x-2) ]. Cancel (x+2): lim_(x→-2) (x² - 2x + 4) / (x-2). Substitute x = -2: ((-2)² - 2*(-2) + 4) / (-2 - 2) = (4 + 4 + 4) / (-4) = 12 / (-4) = -3.

We use the known limit lim_(θ→0) sin(θ)/θ = 1. To match the form, we set θ = 3x. Then as x→0, θ→0. Rewrite the limit: lim_(x→0) sin(3x)/x = lim_(x→0) [ (sin(3x))/(3x) * 3 ]. This equals 3 * lim_(θ→0) sin(θ)/θ (where θ=3x). Since lim_(θ→0) sin(θ)/θ = 1, the limit is 3 * 1 = 3.

Direct substitution gives 0/0. Factor the denominator: x² - 4 = (x-2)(x+2). The limit becomes lim_(x→2) (x-2) / [(x-2)(x+2)]. Cancel the common factor (x-2), provided x ≠ 2, to get lim_(x→2) 1/(x+2). Now substitute x=2: 1/(2+2) = 1/4.