Infinity Limits Mastery Quiz

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1) Evaluate lim_x→∞(3x⁵ + 2x²)/(4x⁵ - 7x³ + 1)

3/4

-3/4

∞

Does not exist

Degrees are equal (5). The leading coefficients are 3 and 4, so the ratio is 3/4. When x is large negative, both leading terms are negative, but negative divided by negative is positive, so the limit is still 3/4.

Explanation

About This Quiz

Ready for a challenge? This quiz takes you into deeper territory, comparing extreme growth rates like factorials versus exponentials, exponentials versus polynomials, and roots versus logarithms. You’ll evaluate complex expressions using conjugates, analyze large-x behavior in inverse trig functions, and determine horizontal asymptotes with precision. Each question helps you understand... see morehow mathematicians classify functions by growth and why limits at infinity are essential in calculus.
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2) Evaluate lim_x→∞ (√(25x² + 20x + 1) - 5x)

1/5

4/5

∞

Multiply and divide by the conjugate: [ (25x² + 20x + 1) - 25x² ] / [ √(25x² + 20x + 1) + 5x ] = (20x + 1) / [ √(25x² + 20x + 1) + 5x ]. Divide numerator and denominator by x (x > 0): (20 + 1/x) / [ √(25 + 20/x + 1/x²) + 5 ]. As x approaches infinity, this simplifies to 20 / (5 + 5) = 20/10 = 2.

Explanation

3) Evaluate lim_x→∞ x⁴ / (3^x)

∞

Exponential functions grow faster than any polynomial. Applying L’Hôpital’s rule four times eventually gives a constant over (ln 3)⁴ 3^x , which goes to 0.

Explanation

Exponential functions grow faster than any polynomial. Applying L’Hôpital’s rule four times eventually gives a constant over (ln 3)⁴ 3^x , which goes to 0.

4) Evaluate lim(x→∞) ln(100x) / x²

∞

100

The logarithm grows much slower than any positive power of x. Here the denominator is x², so ln(100x)/x² → 0 (L’Hôpital gives 1/x / 2x = 1/(2x²) → 0).

Explanation

The logarithm grows much slower than any positive power of x. Here the denominator is x², so ln(100x)/x² → 0 (L’Hôpital gives 1/x / 2x = 1/(2x²) → 0).

5) Evaluate lim_x→∞ (7 - 2/x)/(2 + 5/x³)

7/2

∞

As x → ∞, terms with 1/x and 1/x³ vanish, leaving 7/2 exactly.

Explanation

As x → ∞, terms with 1/x and 1/x³ vanish, leaving 7/2 exactly.

6) Evaluate lim_x→∞ sin(5x)/√x)

Does not exist

|sin(5x)| ≤ 1, so |sin(5x)/√x| ≤ 1/√x → 0. By the squeeze theorem the limit is 0.

Explanation

|sin(5x)| ≤ 1, so |sin(5x)/√x| ≤ 1/√x → 0. By the squeeze theorem the limit is 0.

7) Evaluate lim_x→∞ (x³ + 5x)/(x³ + sin(x))

∞

Does not exist

Divide numerator and denominator by x³: (1 + 5/x²)/(1 + sin(x)/x³). As x → ∞, sin(x)/x³ → 0, so the limit is 1/1 = 1.

Explanation

Divide numerator and denominator by x³: (1 + 5/x²)/(1 + sin(x)/x³). As x → ∞, sin(x)/x³ → 0, so the limit is 1/1 = 1.

8) Evaluate lim_x→∞ e(-x²) · x¹⁰

∞

The Gaussian exponential e^{-x²} goes to 0 faster than any polynomial grows, so the product goes to 0.

Explanation

The Gaussian exponential e^{-x²} goes to 0 faster than any polynomial grows, so the product goes to 0.

9) Evaluate lim_x→∞ [√(x² - 6x) - x]

-3

-6

Multiply by conjugate: (x² - 6x - x²) / [√(x² - 6x) + x] = -6x / [√(x² - 6x) + x]. Divide numerator and denominator by x (x > 0): -6 / [√(1 - 6/x) + 1] → -6 / (1 + 1) = -6/2 = -3.

Explanation

Multiply by conjugate: (x² - 6x - x²) / [√(x² - 6x) + x] = -6x / [√(x² - 6x) + x]. Divide numerator and denominator by x (x > 0): -6 / [√(1 - 6/x) + 1] → -6 / (1 + 1) = -6/2 = -3.

10) Evaluate lim_x→∞ x (√(1 + 7/x) - 1)

7/2

∞

Multiply numerator by conjugate: x [(1 + 7/x - 1)/(√(1 + 7/x) + 1)] = x (7/x) / (√(1 + 7/x) + 1) = 7 / (√(1 + 7/x) + 1) → 7 / (1 + 1) = 7/2.

Explanation

Multiply numerator by conjugate: x [(1 + 7/x - 1)/(√(1 + 7/x) + 1)] = x (7/x) / (√(1 + 7/x) + 1) = 7 / (√(1 + 7/x) + 1) → 7 / (1 + 1) = 7/2.

11) Evaluate lim_x→∞ (tan⁻¹ x) / x

π/2

∞

arctan(x) approaches π/2, a constant, while x → ∞, so constant over infinity is 0.

Explanation

arctan(x) approaches π/2, a constant, while x → ∞, so constant over infinity is 0.

12) Evaluate lim_x→∞ (5x) / (x!)

∞

A factorial grows faster than any exponential with a fixed base (such as 5x). Since the denominator grows much faster than the numerator, the terms eventually decrease to 0. So the limit is 0.

Explanation

A factorial grows faster than any exponential with a fixed base (such as 5x). Since the denominator grows much faster than the numerator, the terms eventually decrease to 0. So the limit is 0.

13) The function f(x) = (2x² - 3x + 1)/(x² + 10) has horizontal asymptote(s):

Y = 0

Y = 2 (both sides)

Y = 2 as x → ∞ and y = -2 as x → -∞

None

Degree 2 over degree 2, ratio of leading coefficients 2/1 = 2, and the sign is positive on both sides.

Explanation

Degree 2 over degree 2, ratio of leading coefficients 2/1 = 2, and the sign is positive on both sides.

14) What is lim_x→∞ e4x / e5x ?

∞

E-x

e4x / e5x = e4x - 5x = e-x which goes to 0 as x → ∞

Explanation

e4x / e5x = e4x - 5x = e-x which goes to 0 as x → ∞

15) Evaluate lim_x→∞ (9x⁶ + 100x²)/( -3x⁶ + x⁵ - 7)

-3

-1/3

Leading coefficients 9 and -3, ratio 9/(-3) = -3.

Explanation

Leading coefficients 9 and -3, ratio 9/(-3) = -3.

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Alva Benedict B. |PhD

College Expert

Alva Benedict B. is an experienced mathematician and math content developer with over 15 years of teaching and tutoring experience across high school, undergraduate, and test prep levels. He specializes in Algebra, Calculus, and Statistics, and holds advanced academic training in Mathematics with extensive expertise in LaTeX-based math content development.