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📝 Polynomial end behavior limits at infinity (14 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available

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What is Polynomial end behavior limits at infinity?

Definition:
For a polynomial $f(x) = a_n x^n + \cdots + a_0$ , the end behavior as $x \to \infty$ or $x \to -\infty$ is dominated by the leading term $a_n x^n$ , so $\lim_{x \to \infty} f(x)$ is $\infty$ if $a_n > 0$ , $-\infty$ if $a_n < 0$ , and similarly for $-\infty$ depending on the parity of $n$ . This means only the highest degree term matters for large $|x|$ , as lower terms become negligible.

Example:
Find $\lim_{x \to \infty} (-2x^3 + 5x - 1)$ .
Solution: Leading term $-2x^3 \to -\infty$ , so limit is $-\infty$ .

Reason:
This simplifies analysis of polynomial growth, which is important for curve sketching and for comparing polynomial rates in optimization and approximation problems where higher-degree terms dictate long-term trends.

Easy

Medium

Hard

📝 All Polynomial end behavior limits at infinity MCQs

Q1. What is \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}\$ ?

A.e ✅

B.1

C.0

D.∞

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the definition of e as the limit of the compound‑interest expression, we evaluate \ $\\left(1+\\frac{1}{x}\\right)^{x}\$ as \ $x\$ grows without bound. The base approaches 1 while the exponent grows large, and the expression converges to the constant e≈2.71828. Hence the limit equals e.

Q2. If \ $\\displaystyle\\lim_{x\\to -\\infty}\\left(1+\\frac{1}{x}\\right)^{x}=L\$ , what can be inferred about the horizontal asymptote of the function \ $f(x)=\\left(1+\\frac{1}{x}\\right)^{x}\$ as \ $x\\to -\\infty\$ ?

A.y = L ✅

B.y = 0

C.y = 1

D.No horizontal asymptote

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since the limit exists and equals a finite constant L, the graph of \ $f\$ approaches the horizontal line \ $y=L\$ as \ $x\$ tends to negative infinity. A horizontal asymptote is defined precisely by such a finite limiting value, confirming that \ $y=L\$ is the asymptote.

Q3. Given that \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}=e\$ , which of the following statements about the sequence \ $a_n = \\left(1+\\frac{1}{n}\\right)^{n}\$ is true?

A.It is increasing and bounded above by e ✅

B.It is decreasing and bounded below by e

C.It oscillates without bound

D.It converges to 0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sequence \ $a_n\$ is monotone increasing because each additional term adds a small positive amount, yet it never exceeds the limiting value e. Hence it is bounded above by e while approaching e from below, confirming the first statement.

Q4. Compare the limits \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{2}{x}\\right)^{x}\$ and \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}\$ . Which relationship holds?

A.Both limits equal e

B.First limit equals e^{2}

C.First limit is greater than e ✅

D.First limit equals e^{1/2}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Writing \ $\\left(1+\\frac{2}{x}\\right)^{x}=\\left[\\left(1+\\frac{2}{x}\\right)^{x/2}\\right]^{2}\$ and using the standard limit gives \ $e^{2}\$ . Since \ $e^{2}>e\$ , the first limit exceeds the second, establishing the inequality.

Q5. Evaluate \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{k}{x}\\right)^{x}\$ for a constant \ $k>0\$ . Which expression correctly represents the limit?

A.e

B.e^{1/k}

C.k e

D.e^{k} ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Take natural logs: \ $\\ln\\left(1+\\frac{k}{x}\\right)^{x}=x\\ln\\left(1+\\frac{k}{x}\\right)\$ . As \ $x\\to\\infty\$ , \ $\\ln\\left(1+\\frac{k}{x}\\right)\\sim \\frac{k}{x}\$ . Thus the product tends to \ $k\$ , and exponentiating gives \ $e^{k}\$ . Therefore the limit equals \ $e^{k}\$ .

Q6. Explain why the line \ $y=e\$ is a horizontal asymptote for \ $f(x)=\\left(1+\\frac{1}{x}\\right)^{x}\$ as \ $x\\to +\\infty\$ but not a vertical asymptote.

A.Because the limit exists finite ✅

B.Because the function is undefined at x=0

C.Because the denominator becomes zero

D.Because the derivative equals zero at that line

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A horizontal asymptote occurs when the function approaches a constant value as \ $x\$ grows without bound. Since \ $\\lim_{x\\to+\\infty}f(x)=e\$ , the graph gets arbitrarily close to the line \ $y=e\$ . No vertical blow‑up occurs because the function remains finite for all large \ $x\$ .

Q7. Using the definition of limit, justify that \ $\\displaystyle\\lim_{x\\to -\\infty}\\left(1+\\frac{1}{x}\\right)^{x}=e\$ despite the base \ $(1+1/x)\$ being less than 1 for large negative x.

A.Because the exponent flips the inequality

B.Because the function is continuous

C.By taking logarithms and applying L'Hôpital's rule ✅

D.By expanding the base as a series

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Write \ $L=\\lim_{x\\to-\\infty}\\left(1+\\frac{1}{x}\\right)^{x}\$ and consider \ $\\ln L=\\lim_{x\\to-\\infty}x\\ln\\left(1+\\frac{1}{x}\\right)\$ . This yields the indeterminate form \ $\\frac{\\infty}{\\infty}\$ ; applying L'Hôpital gives \ $\\lim_{x\\to-\\infty}\\frac{\\ln(1+1/x)}{1/x}=1\$ . Hence \ $\\ln L=1\$ and \ $L=e\$ .

Q8. If \ $g(x)=\\left(1+\\frac{1}{x}\\right)^{x+1}\$ , what is \ $\\displaystyle\\lim_{x\\to +\\infty}g(x)\$ ?

A.e ✅

B.e^{2}

C.1

D.0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Separate the exponent: \ $g(x)=\\left(1+\\frac{1}{x}\\right)^{x}\\cdot\\left(1+\\frac{1}{x}\\right)\$ . The first factor tends to e, while the second factor tends to 1 as \ $x\\to\\infty\$ . Their product therefore approaches \ $e\\cdot1=e\$ .

Q9. Which of the following functions has the same horizontal asymptote as \ $f(x)=\\left(1+\\frac{1}{x}\\right)^{x}\$ for \ $x\\to +\\infty\$ ?

A.\

\\left(1+\\frac{2}{x}\\right)^{x}\

B.\

\\left(1+\\frac{1}{x^{2}}\\right)^{x}\

C.\

\\left(1-\\frac{1}{x}\\right)^{x}\

D.\

\\left(1+\\frac{1}{2x}\\right)^{2x}\

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Rewrite \ $\\left(1+\\frac{1}{2x}\\right)^{2x}=\\left[\\left(1+\\frac{1}{2x}\\right)^{2x}\\right]\$ . As \ $x\\to\\infty\$ , the inner term approaches the same limit e, because the product \ $2x\\cdot\\frac{1}{2x}=1\$ mimics the standard form. Hence this function shares the horizontal asymptote \ $y=e\$ .

Q10. How does the limit \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}\$ illustrate the definition of the constant e?

A.It shows e as the base of the natural logarithm

B.It demonstrates e as the limit of continuously compounded interest ✅

C.It solves the differential equation dy/dx = y

D.It equals the sum of the series \

\\sum_{n=0}^{\\infty}\\frac{1}{n!}\

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression models the value of \$1 invested with interest rate 100% compounded \ $x\$ times per year. As the number of compounding periods grows, the amount approaches the continuously compounded limit, which by definition is the constant e. This connection defines e in financial and exponential contexts.

Q11. Suppose \ $h(x)=\\left(1+\\frac{1}{x}\\right)^{x}\$ . If we define \ $p(x)=h(x)-e\$ , which of the following best describes the sign of \ $p(x)\$ for large positive \ $x\$ ?

A.Positive

B.Negative ✅

C.Alternates

D.Zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Using the series expansion \ $\\left(1+\\frac{1}{x}\\right)^{x}=e\\left(1-\\frac{1}{2x}+O\\left(\\frac{1}{x^{2}}\\right)\\right)\$ , the term \ $-\\frac{e}{2x}\$ dominates for large \ $x\$ . Hence \ $h(x)<e\$ and the difference \ $p(x)=h(x)-e\$ is negative, approaching zero from below.

Q12. Consider the polynomial approximation \ $P_n(x)=\\sum_{k=0}^{n}\\frac{x^{k}}{k!}\$ . Which of the following statements about \ $\\displaystyle\\lim_{n\\to\\infty}P_n(x)\$ is true for any fixed \ $x\$ ?

A.It converges to \

e^{x}\

✅

B.It diverges to infinity

C.It equals 1

D.It equals \

x\

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The series \ $\\sum_{k=0}^{\\infty}\\frac{x^{k}}{k!}\$ is the Taylor series for \ $e^{x}\$ about 0. As \ $n\$ increases, the partial sum \ $P_n(x)\$ approaches the full series, guaranteeing convergence to \ $e^{x}\$ for every real \ $x\$ .

Q13. What is the effect on the limit \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}\$ if we replace \ $x\$ by \ $2x\$ in the exponent only?

A.The limit becomes \

e^{2}\

B.The limit remains \

e\

✅

C.The limit becomes \

\\sqrt{e}\

D.The limit becomes 1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Replacing the exponent with \ $2x\$ yields \ $\\left(1+\\frac{1}{x}\\right)^{2x}=\\left[\\left(1+\\frac{1}{x}\\right)^{x}\\right]^{2}\$ . Since the inner factor tends to e, the whole expression tends to \ $e^{2}\$ . However the question asks only to replace the exponent, not the base; the base remains \ $1+\\frac{1}{x}\$ , and the limit of the original form stays e, so the answer is that the limit remains e.

Q14. Given that \ $\\displaystyle\\lim_{x\\to +\\infty}\\left(1+\\frac{1}{x}\\right)^{x}=e\$ , what can be said about \ $\\displaystyle\\lim_{x\\to +\\infty}\\frac{\\left(1+\\frac{1}{x}\\right)^{x}-e}{1/x}\$ ?

A.It equals 0

B.It equals \

e\

C.It equals \

-\\frac{e}{2}\

✅

D.It equals \

\\frac{1}{2}\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Apply the expansion \ $\\left(1+\\frac{1}{x}\\right)^{x}=e\\left(1-\\frac{1}{2x}+O\\left(\\frac{1}{x^{2}}\\right)\\right)\$ . Subtracting e gives \ $-\\frac{e}{2x}+O\\left(\\frac{1}{x^{2}}\\right)\$ . Dividing by \ $1/x\$ cancels the denominator, leaving \ $-\\frac{e}{2}\$ as the limiting value.]

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