Calculus: Evaluating Limits Algebraically
Using substitution, factoring, rationalizing, and limit laws
Calculus: Evaluating Limits Algebraically
Using substitution, factoring, rationalizing, and limit laws
Math - Grade 9-12
- 1
Evaluate the limit as x approaches 3: 2x^2 - 5x + 1.
The limit is 4. Since this is a polynomial, direct substitution works: 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4. - 2
Evaluate the limit as x approaches 3: (x^2 - 9)/(x - 3).
Factor the difference of squares before substituting.
The limit is 6. Factor the numerator as (x - 3)(x + 3), cancel the common factor x - 3, and then substitute x = 3 into x + 3. - 3
Evaluate the limit as x approaches -2: (x^2 + 5x + 6)/(x + 2).
The limit is 1. Factor the numerator as (x + 2)(x + 3), cancel x + 2, and substitute x = -2 into x + 3 to get 1. - 4
Evaluate the limit as x approaches 4: (sqrt(x + 5) - 3)/(x - 4).
Use the conjugate sqrt(x + 5) + 3 to simplify the numerator.
The limit is 1/6. Multiply by the conjugate to get 1/(sqrt(x + 5) + 3), then substitute x = 4 to get 1/(3 + 3) = 1/6. - 5
Evaluate the limit as x approaches 5: (1/x - 1/5)/(x - 5).
Combine the fractions in the numerator first.
The limit is -1/25. Rewrite the numerator as (5 - x)/(5x), so the expression becomes (5 - x)/(5x(x - 5)) = -1/(5x). Substituting x = 5 gives -1/25. - 6
Evaluate the limit as x approaches 2: (x^3 - 8)/(x - 2).
The limit is 12. Factor the numerator as (x - 2)(x^2 + 2x + 4), cancel x - 2, and substitute x = 2 to get 4 + 4 + 4 = 12. - 7
Evaluate the limit as x approaches 4: (x^2 - 4x)/(x^2 - 16).
The limit is 1/2. Factor the expression as x(x - 4)/((x - 4)(x + 4)), cancel x - 4, and substitute x = 4 to get 4/8 = 1/2. - 8
Evaluate the limit as h approaches 0: ((3 + h)^2 - 9)/h.
Expand (3 + h)^2 before simplifying.
The limit is 6. Expand the numerator to get 9 + 6h + h^2 - 9 = 6h + h^2, then divide by h to get 6 + h. Substituting h = 0 gives 6. - 9
Evaluate the limit as x approaches 0: sin(5x)/x.
Use the standard limit sin(u)/u = 1 as u approaches 0.
The limit is 5. Rewrite the expression as 5 times sin(5x)/(5x). Since sin(u)/u approaches 1 as u approaches 0, the limit is 5. - 10
Evaluate the limit as x approaches 0: (1 - cos x)/x.
The limit is 0. Multiplying by the conjugate gives sin^2 x divided by x(1 + cos x), which can be written as (sin x/x)(sin x/(1 + cos x)). This approaches 1 times 0/2, so the limit is 0. - 11
Evaluate the limit as x approaches infinity: (5x^2 - 3x + 1)/(2x^2 + 7).
Compare the leading terms of the numerator and denominator.
The limit is 5/2. The numerator and denominator have the same highest power, x^2, so the limit is the ratio of the leading coefficients, 5/2. - 12
Evaluate the limit as x approaches infinity: (4x^3 + x)/(2x^4 - 1).
The limit is 0. The denominator has a higher degree than the numerator, so the denominator grows faster and the fraction approaches 0. - 13
Let f(x) = x^2 + 2 for x < 1 and f(x) = 4x - 1 for x >= 1. Evaluate the limit of f(x) as x approaches 1.
The limit is 3. The left-hand limit is 1^2 + 2 = 3, and the right-hand limit is 4(1) - 1 = 3. Since both one-sided limits are equal, the limit exists and equals 3. - 14
Evaluate the limit as x approaches 2: |x - 2|/(x - 2).
Check the left-hand and right-hand limits separately.
The limit does not exist. For x values less than 2, the expression equals -1, and for x values greater than 2, the expression equals 1. The one-sided limits are not equal. - 15
Evaluate the limit as x approaches 0: (sqrt(1 + x) - 1)/x.
Use the conjugate sqrt(1 + x) + 1.
The limit is 1/2. Multiply by the conjugate to get 1/(sqrt(1 + x) + 1), then substitute x = 0 to get 1/(1 + 1) = 1/2.