Questions: Limits at Infinity

Question 1

What is lim(x→∞) (4x² + 3x) / (2x² − 5)?

• A. ∞, because the numerator grows without bound as x increases
• B. 2, because the ratio of the leading coefficients is 4/2
• C. 0, because the denominator eventually dominates any numerator
• D. 3/(-5) = −3/5, from the ratio of the non-leading constant terms

Answer: (B)

Dividing numerator and denominator by x² gives (4 + 3/x) / (2 − 5/x²). As x→∞, both 3/x and 5/x² vanish to 0, leaving 4/2 = 2. The degree-comparison shortcut: when numerator and denominator have equal degree, the limit at infinity equals the ratio of leading coefficients. Option A is wrong because although the numerator grows, so does the denominator at the same rate. Option D substitutes constants as if x=0, which is invalid.

Question 2

A student says: 'lim(x→∞) f(x) = ∞ and lim(x→0⁺) g(x) = ∞ both involve infinity, so they describe the same type of behavior.' What error is the student making?

• A. No error — both expressions describe a function growing without bound, which is the same phenomenon
• B. They are confusing a limit at infinity (end behavior as x → ∞) with an infinite limit at a finite point (a vertical asymptote near x = 0)
• C. They are confusing horizontal asymptotes with limits that fail to exist
• D. They are treating ∞ as a real number that can be substituted into functions

Answer: (B)

These are fundamentally different phenomena. lim(x→∞) f(x) = ∞ means f grows without bound as x runs to infinity — this describes end behavior along the x-axis, and y = ∞ is not a horizontal asymptote (the function has none). lim(x→0⁺) g(x) = ∞ describes a vertical asymptote near x = 0 — f blows up as you approach a specific finite x-value. The notation looks similar, but the geometric meaning and analysis techniques are entirely different.

Question 3

Most rational function has exactly one horizontal asymptote.

• T. True
• F. False

Answer: False

A rational function has a horizontal asymptote only when the degree of the numerator is less than or equal to the degree of the denominator. If the numerator has higher degree, the function grows without bound — no horizontal asymptote. Furthermore, non-rational functions like f(x) = x / √(x² + 1) can have two different horizontal asymptotes (y = 1 as x → +∞ and y = −1 as x → −∞), because √(x²) = |x| behaves differently in the two directions.

Question 4

A function can cross its horizontal asymptote for some finite value of x, even though the asymptote describes the function's behavior only as x → ∞.

• T. True
• F. False

Answer: True

A horizontal asymptote describes limit behavior — what the function approaches far out along the x-axis. It says nothing about what happens at finite x. For example, f(x) = sin(x)/x has a horizontal asymptote y = 0 as x → ±∞, but it crosses y = 0 infinitely many times at every x = nπ. This is a common misconception: asymptotes describe end behavior, not a boundary the function can never reach.

Question 5

Explain why lim(x→∞) (3x² + 1) / (x² − 5) = 3 rather than ∞/∞, and describe the technique that makes this clear.

Model answer: Substituting ∞ directly would give the indeterminate form ∞/∞, which is undefined. The standard technique is to divide every term in the numerator and denominator by the highest power of x in the denominator — here, x². This gives (3 + 1/x²) / (1 − 5/x²). As x → ∞, both 1/x² and 5/x² approach 0, leaving 3/1 = 3. The technique works because it converts a ratio of growing polynomials into a ratio of terms with known limits.

The key insight is that infinity cannot be substituted as a number — doing so produces ∞/∞, which is indeterminate, not a ratio. Dividing by x² first reveals the true limiting behavior by isolating the leading terms. This same 'divide by highest power' strategy applies to all rational functions and extends to functions with radicals, where the technique requires care about the sign of √(x²) = |x|.