Infinity Limits: Mastery Level

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.

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.

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.

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).

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

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

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.

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

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.

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.

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

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.

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

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

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