Advanced Limits at Infinity Quiz

1) Evaluate lim(x→∞) (7x⁵ + 2x - 1)/(3x⁵ - 8x³ + 5)

0

7/3

1

∞

The degrees of numerator and denominator are both 5. Divide numerator and denominator by x⁵: (7 + 2/x^4 - 1/x⁵)/(3 - 8/x² + 5/x⁵). As x → ∞ all terms with x in the denominator go to 0, leaving 7/3.

Explanation

The degrees of numerator and denominator are both 5. Divide numerator and denominator by x⁵: (7 + 2/x^4 - 1/x⁵)/(3 - 8/x² + 5/x⁵). As x → ∞ all terms with x in the denominator go to 0, leaving 7/3.

2) Evaluate limx→-∞ (4x³ - 9x)/(5x³ + 2)

4/5

-4/5

∞

0

The degrees of the numerator and denominator are equal (both are 3). For rational functions where the degrees are equal, the limit as x approaches positive or negative infinity is simply the ratio of the leading coefficients. The leading coefficients are 4 and 5, so the limit is 4/5.

Explanation

The degrees of the numerator and denominator are equal (both are 3). For rational functions where the degrees are equal, the limit as x approaches positive or negative infinity is simply the ratio of the leading coefficients. The leading coefficients are 4 and 5, so the limit is 4/5.

3) Evaluate limx→∞ √(x² + 5x + 1) - x

0

5/2

1/2

∞

∞ - ∞ form. Multiply by conjugate: [ (x² + 5x + 1) - x² ] / [ √(x² + 5x + 1) + x ] = (5x + 1) / [ √(x² + 5x + 1) + x ]. Divide numerator and denominator by x (x > 0): (5 + 1/x) / [ √(1 + 5/x + 1/x²) + 1 ] → 5 / (1 + 1) = 5/2.

Explanation

∞ - ∞ form. Multiply by conjugate: [ (x² + 5x + 1) - x² ] / [ √(x² + 5x + 1) + x ] = (5x + 1) / [ √(x² + 5x + 1) + x ]. Divide numerator and denominator by x (x > 0): (5 + 1/x) / [ √(1 + 5/x + 1/x²) + 1 ] → 5 / (1 + 1) = 5/2.

4) True or False: If the limit as x approaches infinity of f(x) equals a finite number L, then the line y = L is a horizontal asymptote of the graph of f(x).

True

False

This is the definition of a horizontal asymptote. If the function values f(x) get arbitrarily close to a number L as x increases without bound, then the graph of the function approaches the horizontal line y = L.

Explanation

This is the definition of a horizontal asymptote. If the function values f(x) get arbitrarily close to a number L as x increases without bound, then the graph of the function approaches the horizontal line y = L.

5) Evaluate lim(x→∞) ln(3x + 7) / √x

0

∞

3

7

Logarithm grows slower than any positive power of x. √x = x^{1/2}, so ln(something linear) / x^{1/2} → 0.

Explanation

Logarithm grows slower than any positive power of x. √x = x^{1/2}, so ln(something linear) / x^{1/2} → 0.

6) Evaluate lim(x→-∞) (5x^4 + 3x²)/(-2x^4 + x)

-5/2

5/2

-2/5

0

Highest degree 4 in both. Leading coefficients 5 and -2, ratio 5/(-2) = -5/2.

Explanation

Highest degree 4 in both. Leading coefficients 5 and -2, ratio 5/(-2) = -5/2.

7) Evaluate lim(x→∞) (4 + sin(x))/x

0

4

∞

Does not exist

|sin(x)| ≤ 1, so |(4 + sin(x))/x| ≤ (4 + 1)/|x| = 5/|x| → 0 as x → ∞. By squeeze theorem the limit is 0.

Explanation

|sin(x)| ≤ 1, so |(4 + sin(x))/x| ≤ (4 + 1)/|x| = 5/|x| → 0 as x → ∞. By squeeze theorem the limit is 0.

8) Evaluate lim_x→∞ (x + 5)/(x + sin(x))

0

1

∞

Does not exist

Divide numerator and denominator by x: (1 + 5/x)/(1 + sin(x)/x). sin(x)/x → 0, 5/x → 0, so 1/1 = 1.

Explanation

Divide numerator and denominator by x: (1 + 5/x)/(1 + sin(x)/x). sin(x)/x → 0, 5/x → 0, so 1/1 = 1.

9) Evaluate lim_x→∞ (2x² + 3x)/(e^x + e^{-x})

0

2

∞

1

Denominator ≈ e^x for large positive x, so behaves like (2x²)/e^x → 0.

Explanation

Denominator ≈ e^x for large positive x, so behaves like (2x²)/e^x → 0.

10) Evaluate lim_x→∞ [√(4x² - x) - 2x]

-1/4

0

1/4

-1/2

Conjugate method: multiply by √(4x² - x) + 2x. Numerator (4x² - x) - 4x² = -x. Denominator x (√(4 - 1/x) + 2). Divide num/den by x: -1 / (√(4 - 1/x) + 2) → -1/(2 + 2) = -1/4.

Explanation

Conjugate method: multiply by √(4x² - x) + 2x. Numerator (4x² - x) - 4x² = -x. Denominator x (√(4 - 1/x) + 2). Divide num/den by x: -1 / (√(4 - 1/x) + 2) → -1/(2 + 2) = -1/4.