📝 Infinite limits at infinity (13 MCQs)
📖 From Calculus • 2. Limits and Continuity an Introduction • 13 questions available
What is Infinite limits at infinity?
Definition:
An infinite limit at infinity, denoted or , occurs when the function values grow without bound (positively or negatively) as becomes arbitrarily large. This indicates that the function increases or decreases without approaching any finite horizontal asymptote, often due to polynomial terms with positive leading coefficients or exponential growth.
Example:
Evaluate .
Solution: As , , so .
Reason:
This concept distinguishes functions that grow indefinitely from those that stabilize, which is vital for comparing growth rates in computer science (algorithm complexity), biology (population explosion), and finance (compound interest).
📝 All Infinite limits at infinity MCQs
Q1. Which of the following best describes the meaning of \\\lim_{x\\to +\\infty} f(x)=+\\infty\?
📖 Explanation: The notation means that as x becomes larger and larger, the values of f(x) increase without any upper bound. In other words, for every large number M there exists a number N such that x>N implies f(x)>M. It does not indicate convergence to a finite value, nor does it describe oscillation. Hence option A captures the definition.
Q2. For the function \g(x)=\\frac{2x^{3}-7}{x^{2}+1}\, what is \\\lim_{x\\to -\\infty} g(x)\?
📖 Explanation: The dominant term in the numerator is 2x³ and the denominator behaves like x². As x→-∞, the fraction behaves like 2x³/x² = 2x, which tends to -∞ because x is negative large. Therefore the limit is negative infinity, matching option B.
Q3. Consider \p(x)=\\frac{4x^{5}+3}{-2x^{5}+5x}\. What is \\\lim_{x\\to +\\infty} p(x)\?
📖 Explanation: Both numerator and denominator are degree‑five polynomials. The leading coefficients are 4 and -2, so the fraction approaches the constant ratio 4/(-2)= -2. Since the limit is a finite number, it is not an infinite limit. Option C correctly states that the limit is a finite non‑zero number.
Q4. Evaluate \\\lim_{x\\to +\\infty}\\big(\\sqrt{x^{2}+3x}-x\\big)\.
📖 Explanation: Rewrite \\\sqrt{x^{2}+3x}-x = x\\big(\\sqrt{1+\\frac{3}{x}}-1\\big)\. Using the expansion \\\sqrt{1+u}\\approx1+\\frac{u}{2}\ for small u, we obtain \x\\big(\\frac{3}{2x}\\big)=\\frac{3}{2}\. Hence the expression approaches the finite value 3/2, not an infinite quantity. Option C gives the correct limit.
Q5. Which function grows faster as \x\\to +\\infty\: \f(x)=x^{3}\ or \g(x)=e^{x}\?
📖 Explanation: Polynomial growth of degree three is much slower than exponential growth. As x becomes large, e^{x} dominates any power of x, causing the ratio e^{x}/x³ to diverge to infinity. Therefore e^{x} grows faster than x³, which is described by option C.
Q6. Given \f(x)=\\frac{1}{x}\ and \g(x)=\\frac{1}{x^{2}}\, which statement about their limits as \x\\to +\\infty\ is correct?
📖 Explanation: Both functions approach zero as x→+∞, but the rate differs. Since \\\frac{1}{x^{2}}=(\\frac{1}{x})^{2}\, it decreases quadratically while \\\frac{1}{x}\ decreases linearly. Consequently, \\\frac{1}{x^{2}}\ approaches zero more rapidly than \\\frac{1}{x}\. Option D correctly reflects this relationship.
Q7. Compare the limits of \h(x)=\\ln(x)\ and \k(x)=\\sqrt{x}\ as \x\\to +\\infty\.
📖 Explanation: For large x, \\\sqrt{x}\ behaves like x^{1/2}, while \\\ln x\ grows slower than any positive power of x. Thus both limits are +∞, but the square‑root function increases at a faster rate than the logarithm. Option B correctly states this comparison.
Q8. Why does a rational function whose numerator degree exceeds the denominator by one lead to an infinite limit with sign determined by the leading coefficients?
📖 Explanation: When the degree difference is exactly one, the function behaves like the ratio of leading coefficients multiplied by x. As |x|→∞, the magnitude grows without bound, and the sign follows the sign of that coefficient ratio. Option B accurately captures this principle.
Q9. If \\\lim_{x\\to -\\infty} f(x)= -\\infty\, which statement must be true about the graph of \f\ for large negative \x\?
📖 Explanation: If the limit as x approaches negative infinity is -∞, the function values become arbitrarily large negative numbers. Hence, for sufficiently large negative x, the graph lies below the x‑axis and descends without bound. Option B correctly describes this behavior.
Q10. Which of the following functions does NOT have an infinite limit at either \+\\infty\ or \-\\infty\?
📖 Explanation: The function 1/x tends to zero as x approaches either +∞ or -∞, so it does not diverge to infinity in either direction. All other listed functions have unbounded growth in at least one direction. Therefore option B is the only function without an infinite limit at infinity.
Q11. For the piecewise function \f(x)=\\begin{cases}x^{2}&x\\le 0\\\\ \\frac{1}{x}&x>0\\end{cases}\, what are \\\lim_{x\\to -\\infty} f(x)\ and \\\lim_{x\\to +\\infty} f(x)\?
📖 Explanation: For x→-∞ the piece uses the branch x², which grows like x² and thus tends to +∞. For x→+∞ the branch is 1/x, which approaches zero. Hence the pair of limits is (+∞, 0), corresponding to option A.
Q12. Consider \f(x)=\\frac{x^{3}-6x}{x^{2}-4}\. What is \\\lim_{x\\to -\\infty} f(x)\?
📖 Explanation: The highest‑degree terms dominate: the numerator behaves like x³ and the denominator like x², giving an overall behavior similar to x. As x→-∞, this linear term goes to -\\infty, so the limit is negative infinity. Option A correctly states this.
Q13. Suppose \\\lim_{x\\to +\\infty} f(x)=+\\infty\ and \\\lim_{x\\to -\\infty} f(x)=+\\infty\. Which statement must be true?
📖 Explanation: Both one‑sided limits being +∞ guarantee that beyond some large magnitude R, the function’s values exceed any prescribed positive bound, implying f(x) is positive for all |x|>R. This necessity is expressed in option A. The other statements are not forced by the given limits.