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📝 End behavior and limits of trig functions (13 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 13 questions available

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What is End behavior and limits of trig functions?

Definition:
Trigonometric functions like $\sin x$ and $\cos x$ do not have limits as $x \to \infty$ because they oscillate indefinitely between $-1$ and $1$ , never settling to a single value. However, limits at finite points are well-defined due to continuity, and certain combinations (e.g., $\frac{\sin x}{x}$ ) may have limits at infinity (here $0$ ), but the basic sine and cosine are oscillatory with no end-behavior limit.

Example:
Determine if $\lim_{x \to \infty} \sin x$ exists.
Solution: No, because $\sin x$ oscillates between $-1$ and $1$ , so the limit does not exist.

Reason:
This distinction is important for modeling periodic phenomena like sound waves or alternating current, where the lack of a limit at infinity reflects ongoing oscillation, unlike decaying or stabilizing processes.

Easy

Medium

Hard

📝 All End behavior and limits of trig functions MCQs

Q1. Why does the limit \ $\\lim_{x\\to\\infty}\\sin x\$ fail to exist?

A.Because \

\\sin x\

is unbounded

B.Because \

\\sin x\

oscillates between -1 and 1 ✅

C.Because \

\\sin x\

approaches 0

D.Because \

\\sin x\

has a vertical asymptote

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The sine function never settles to a single value because as x grows it repeatedly returns to every value between -1 and 1. Since the definition of a limit at infinity requires the function to approach one specific real number, the perpetual oscillation prevents the limit from existing.

Q2. Given \ $f(x)=\\sin x+\\frac{1}{x}\$ , what is \ $\\lim_{x\\to\\infty}f(x)\$ ?

A.0

B.1

C.Does not exist ✅

D.-1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The term 1/x tends to 0, but the sine term continues to oscillate between -1 and 1. Adding a vanishing term does not suppress the oscillation, so the combined function does not approach any single number. Therefore the limit does not exist.

Q3. If \ $\\lim_{x\\to\\infty}h(x)=\\infty\$ , what can be said about \ $\\lim_{x\\to\\infty}\\sin h(x)\$ ?

A.The limit does not exist ✅

B.The limit equals 0

C.The limit equals 1

D.The limit equals -1

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the argument of sine grows without bound, the sine values keep cycling through their full range. No matter how large x becomes, the sine function will still produce values arbitrarily close to any number between -1 and 1, so the limit cannot be defined.

Q4. Why does \ $\\lim_{x\\to\\infty}\\sin(x^2)\$ not exist?

A.Because \

\\sin\

is unbounded

B.Because \

x^2\

grows without bound causing the argument to oscillate infinitely

C.Because \

\\sin\

approaches 0 as its argument grows

D.Because the limit of the polynomial is infinite ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The polynomial x^2 increases to infinity, causing the argument of the sine function to sweep through infinitely many periods. This continual passage through all phases of the sine wave means the function never settles to a single value, so the limit at infinity does not exist.

Q5. Which statement correctly compares the end behavior of \ $y=\\sin x\$ and \ $y=\\cos x\$ as \ $x\\to\\infty\$ ?

A.Both tend to 0

B.Both diverge to infinity

C.Both fail to have limits because of periodicity ✅

D.Sin has a limit but cos does not

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Both sine and cosine are periodic with period 2π and are bounded between -1 and 1. Because they repeat their values indefinitely as x grows, neither function approaches a particular number. Thus they share the same type of end behavior: no limit exists due to periodicity.

Q6. Which of the following functions possesses a horizontal asymptote at \ $y=0\$ ?

A.\

\\frac{\\sin x}{x}\

✅

B.\

\\sin x\

C.\

\\tan x\

D.\

\\sec x\

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The quotient sin x / x is bounded by 1/|x| in absolute value, and as x approaches infinity the denominator grows without bound, forcing the whole expression toward zero. This satisfies the definition of a horizontal asymptote at y = 0, unlike the other listed functions.

Q7. Given \ $f(x)=\\sin x\$ and \ $g(x)=\\sin(2x)\$ , which statement about their end behavior is true?

A.Both have the same limit at infinity

B.g(x) oscillates twice as fast but still has no limit ✅

C.f(x) has a limit while g(x) does not

D.g(x) has limit 0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Doubling the argument speeds up the oscillation, but the sine function still only takes values between -1 and 1 and never settles. Both functions lack a limit at infinity, and the faster oscillation of sin(2x) does not create a limit. Hence the correct statement is that g(x) oscillates twice as fast but still has no limit.

Q8. Consider \ $\\lim_{x\\to\\infty}\\sin\\left(\\frac{1}{x}\\right)\$ and \ $\\lim_{x\\to\\infty}\\frac{\\sin x}{x}\$ . Which choice correctly describes both limits?

A.Both equal 0 ✅

B.First equals 0, second does not exist

C.Both do not exist

D.First does not exist, second equals 0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: As x grows, 1/x approaches 0, making sin(1/x) approach sin(0)=0. Meanwhile, sin x remains bounded while the denominator x grows, forcing sin x / x toward 0 by the squeeze theorem. Both limits therefore equal 0.

Q9. Consider the sequence of functions \ $s_n(x)=\\sin(nx)\$ . For a fixed \ $x\$ that is not a multiple of \ $\\pi\$ , which statement about \ $\\lim_{n\\to\\infty}s_n(x)\$ is correct?

A.The limit exists and equals 0

B.The limit exists and equals \

\\sin x\

C.The limit does not exist ✅

D.The limit exists and equals 1

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a fixed x that is not a multiple of π, the sequence sin(nx) samples the sine function at points that become dense in the interval [‑π,π] as n increases. Consequently the values keep jumping among many different numbers, preventing convergence. Hence the limit does not exist.

Q10. If a function is periodic with period \ $2\\pi\$ and bounded, which statement about its limit as \ $x\\to\\infty\$ is necessarily true?

A.It does not exist unless the function is constant ✅

B.It always exists

C.It equals the average value over a period

D.It equals 0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A bounded periodic function repeats its pattern indefinitely. Unless the function is constant, its values keep cycling and never settle to a single number as x grows. Therefore the limit at infinity does not exist unless the function reduces to a constant function.

Q11. For the composite function \ $q(x)=\\sin(e^x)\$ , which description best captures its end behavior as \ $x\\to\\infty\$ ?

A.Approaches a finite limit

B.Grows without bound

C.Approaches 0

D.Oscillates increasingly rapidly without limit ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The exponential e^x grows extremely fast, causing its argument inside the sine to increase without bound. The sine of an ever‑increasing argument continues to oscillate between –1 and 1, and the oscillations become more rapid. No single value is approached, so the limit does not exist.

Q12. Which statement is true about the limit of a bounded periodic function as \ $x\\to\\infty\$ ?

A.It always exists

B.It never exists

C.It does not exist unless the function is constant ✅

D.It equals the average value over a period

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A bounded periodic function repeats the same finite set of values over each period. Because it never stabilizes to one value, the limit as x approaches infinity cannot exist unless the function is constant. Thus the correct statement is that the limit does not exist unless the function is constant.

Q13. What is the period of the function \ $y=\\sin x\$ ?

A.\

\\pi\

B.\

2\\pi\

✅

C.\

\\frac{\\pi}{2}\

D.\

4\\pi\

💡 Difficulty: easy | ✅ Correct: B

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