📝 Limits of polynomial functions (15 MCQs)
📖 From Calculus • 2. Limits and Continuity an Introduction • 15 questions available
What is Limits of polynomial functions?
Definition:
The limit of a polynomial function as is simply the value of the polynomial at , i.e., , because polynomials are continuous everywhere. This result follows from applying the sum, product, and constant multiple limit laws repeatedly to the basic limit , making evaluation straightforward.
Example:
Evaluate .
Solution: Substitute : .
Reason:
This property ensures that polynomial limits are always computable by direct substitution, providing a simple entry point for students to practice limits before tackling more complex functions like rational or piecewise functions.
📝 All Limits of polynomial functions MCQs
Q1. According to Theorem 1.2.3, the limit of a polynomial p(x) as x→a is …
📖 Explanation: Theorem 1.2.3 states that a polynomial is continuous everywhere, so the limit as x approaches a simply equals the polynomial’s value at a. This follows from the fact that limits of sums, constants, and powers reduce to the limits of the individual terms, which are just the coefficients evaluated at a.
Q2. Which of the following expressions is NOT a polynomial?
📖 Explanation: A polynomial must have non‑negative integer exponents and rational coefficients. The expression √x + 1 contains a fractional exponent (½), violating the definition of a polynomial, whereas the other choices consist solely of integer powers and constant coefficients.
Q3. If limₓ→a p(x)=p(a) holds for every polynomial p, what property does this guarantee for polynomials at x=a?
📖 Explanation: The equality of the limit and the function value at a means that the function does not jump, break, or become undefined there. This is precisely the definition of continuity at a point, so every polynomial is continuous for all real a.
Q4. Let p(x) be a polynomial and k a constant. If limₓ→a p(x)=L, what is limₓ→a (p(x)+k)?
📖 Explanation: Adding a constant k does not affect the limiting process except by shifting the result. Since limits preserve addition, the limit of p(x)+k is the limit of p(x) plus the limit of the constant k, which is k itself, giving L + k.
Q5. Suppose p(x) is a polynomial with limₓ→2 p(x)=0. Define r(x)=(x‑2)p(x). What is limₓ→2 r(x)?
📖 Explanation: Both factors in r(x) approach 0 as x→2: the term (x‑2) tends to 0, and by hypothesis p(x) also tends to 0. The product of two quantities that each approach 0 must itself approach 0, so the limit of r(x) is 0.
Q6. Consider p(x)=x²‑4x+3 and q(x)=2x‑5. Which statement about limₓ→5 p(x)/q(x) is true?
📖 Explanation: Evaluating directly gives p(5)=25‑20+3=8 and q(5)=10‑5=5. Since the denominator is non‑zero at x=5, the limit of the quotient equals the quotient of the limits, which is 8/5. No indeterminate form appears.
Q7. Compute limₓ→2 (x³‑8)/(x‑2).
📖 Explanation: Factor the numerator as (x‑2)(x²+2x+4). Cancelling the common factor (x‑2) leaves x²+2x+4, whose value at x=2 is 4+4+4=12. Hence the limit equals 12, confirming the removable discontinuity at x=2.
Q8. Evaluate limₓ→1 (x⁷‑2x⁵+1)/35.
📖 Explanation: Plugging x=1 into the numerator gives 1‑2+1=0. The denominator 35 is constant, so the whole expression approaches 0/35=0. Because the polynomial numerator is continuous, direct substitution yields the limit.
Q9. Find limₓ→2 (x⁴‑16)/(x²‑4).
📖 Explanation: Factor x⁴‑16 as (x²‑4)(x²+4). Cancel the common factor (x²‑4) to obtain x²+4. Substituting x=2 gives 4+4=8, so the limit is 8. The original expression had a removable discontinuity that disappears after simplification.
Q10. Using Theorem 1.2.3, what is limₓ→‑1 (3x⁵‑2x+7)?
📖 Explanation: Apply the theorem by evaluating the polynomial at x=‑1: 3(‑1)⁵=‑3, ‑2(‑1)=+2, and +7, giving ‑3+2+7=6. Because polynomials are continuous, the limit equals the function value, which is 6.
Q11. Why does evaluating a polynomial at a point give the same result as taking its limit as x approaches that point?
📖 Explanation: A polynomial can be written as a sum of monomials, each of which is a constant times xⁿ. The limit of a constant is the constant itself, and the limit of xⁿ as x→a is aⁿ. Adding these limits reproduces the original polynomial evaluated at a, establishing equality.
Q12. Let p(x) have a root at x=a (i.e., p(a)=0). For r(x)=p(x)/(x‑a), what is limₓ→a r(x) equal to?
📖 Explanation: Because p(x) is divisible by (x‑a) by the Factor Theorem, we can write p(x)=(x‑a)·q(x) where q(x) is a polynomial. Then r(x)=q(x), and the limit as x→a is simply q(a)=p′(a) by the definition of the derivative of p at a.
Q13. Determine limₓ→3 (x³‑27)/(x‑3) and explain its connection to the derivative of f(x)=x³ at x=3.
📖 Explanation: Factor the numerator as (x‑3)(x²+3x+9) and cancel (x‑3), leaving x²+3x+9. Substituting x=3 gives 9+9+9=27? Wait, correct calculation: 3²+3·3+9=9+9+9=27, but the limit of the original quotient equals the derivative f′(3)=3·3²=27. However the simplified expression yields 27, so the correct answer is 27. (Correct option C)
Q14. Given p(x)=x³‑6x²+11x‑6, factor it and compute limₓ→1 p(x)/(x‑1).
📖 Explanation: Factor p(x) as (x‑1)(x‑2)(x‑3). Cancelling (x‑1) leaves (x‑2)(x‑3). Substituting x=1 gives (‑1)(‑2)=2. Hence the limit equals 2, demonstrating how factoring removes the removable discontinuity.
Q15. Which theorem guarantees that limₓ→a p(x)=p(a) for any polynomial p?
📖 Explanation: Theorem 1.2.3 explicitly states that the limit of a polynomial at any real number a coincides with the polynomial’s value at a. This result follows from the continuity of polynomials and is the foundational tool used throughout the chapter on limits of polynomial functions.