Algebraic Limits: Core Concepts and Techniques

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Quizzes Created: 8157 | Total Attempts: 9,569,759
| Attempts: 11 | Questions: 15 | Updated: Dec 17, 2025
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1) Evaluate limx→5 (2x² - 3x + 1)

Explanation

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.

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About This Quiz
Algebraic Limits: Core Concepts And Techniques - Quiz

Are you ready to explore how algebra can make tricky limits easier to understand? In this quiz, you will practice evaluating limits by factoring expressions, canceling common terms, and using conjugates to simplify radical expressions. You will also work with known limit patterns, apply limit laws, and learn how to... see morehandle indeterminate forms by rewriting expressions in simpler ways. Step by step, you will build confidence in using algebraic techniques to make even complicated limits feel manageable and straightforward.
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2) Evaluate lim_(x→1) (x³ - 1) / (x² - 1)

Explanation

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.

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3) Evaluate limx→3 (x² - 9) / (x - 3)

Explanation

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.

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4) Find limx→4 (√x)/(x - 4)

Explanation

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.

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5) Evaluate limx→0 ( (x + 2)² - 4 ) / x

Explanation

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.

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6) Evaluate limx→16 (-4+√x) / (x - 16)

Explanation

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.

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7) If lim_(x→2) f(x) = 5 and lim_(x→2) g(x) = -3, find lim_(x→2) [ 2f(x) - g(x) ].

Explanation

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.

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8) Evaluate lim_(x→-1) (x² + 3x + 2) / (x³ + 1)

Explanation

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.

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9) Evaluate lim_(x→0) [ 1/(x+2) - 1/2 ] / x

Explanation

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.

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10) Evaluate lim_(x→∞) (3x² - 2x + 5) / (4x² + x + 1)

Explanation

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).

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11) Find lim_(x→3) ( |x - 3| ) / (x - 3)

Explanation

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

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12) Given that lim_(x→c) f(x) = 8 and lim_(x→c) g(x) = 2, find lim_(x→c) [ f(x) / g(x) ].

Explanation

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.

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13) Evaluate lim_(x→-2) (x³ + 8) / (x² - 4)

Explanation

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.

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14) Evaluate lim_(x→0) sin(3x) / x. (Hint: You may use the known limit lim_(θ→0) sin(θ)/θ = 1)

Explanation

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.

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15) Evaluate lim_(x→2) (x - 2) / (x² - 4)

Explanation

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.

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Evaluate limx→5 (2x² - 3x + 1)
Evaluate lim_(x→1) (x³ - 1) / (x² - 1)
Evaluate limx→3 (x² - 9) / (x - 3)
Find limx→4 (√x)/(x - 4)
Evaluate limx→0 ( (x + 2)² - 4 ) / x
Evaluate limx→16 (-4+√x) / (x - 16)
If lim_(x→2) f(x) = 5 and lim_(x→2) g(x) = -3, find...
Evaluate lim_(x→-1) (x² + 3x + 2) / (x³ + 1)
Evaluate lim_(x→0) [ 1/(x+2) - 1/2 ] / x
Evaluate lim_(x→∞) (3x² - 2x + 5) / (4x² + x + 1)
Find lim_(x→3) ( |x - 3| ) / (x - 3)
Given that lim_(x→c) f(x) = 8 and lim_(x→c) g(x) = 2, find...
Evaluate lim_(x→-2) (x³ + 8) / (x² - 4)
Evaluate lim_(x→0) sin(3x) / x. (Hint: You may use the known limit...
Evaluate lim_(x→2) (x - 2) / (x² - 4)
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