Limit and Continuity

Maths, HSSC-Punjab First Year. This chapter is split into 37 topics and has 1,002 practice questions. The topics covered are listed below, followed by 10 free sample questions from the chapter.

Topics in this chapter

1. Numerical approach to limits tables of values 21 questions
2. Theorems on limits statements and proofs 30 questions
3. Continuity of a function at a point conditions 30 questions
4. Applications of limits economics astronomy and signals 30 questions
5. Rationalising limits other forms 25 questions
6. Limits at infinity with square roots and the sign of x 30 questions
7. Zero over zero form cubic and higher polynomial limits 26 questions
8. Concept of limit of a function circumscribed polygon 21 questions
9. Limit of ratio of powers xᵐ minus aᵐ over xⁿ minus aⁿ 25 questions
10. One-sided limits left hand and right hand 30 questions
11. Trigonometric limits of sine and tangent ratios 30 questions
12. The limit of (1 + x) to the power 1/x as x tends to zero 30 questions
13. The number e as a limit of (1 + 1/n)ⁿ 29 questions
14. Evaluating limits by direct substitution polynomials 20 questions
15. Applications of limits growth and decay 26 questions
16. Direct substitution rational and radical functions 24 questions
17. Trigonometric limits involving one minus cosine 30 questions
18. Limits at infinity of reciprocal powers 30 questions
19. Continuity of piecewise functions and jump discontinuity 30 questions
20. Zero over zero form factorisation of quadratics 25 questions
21. Limit of sequences extra forms 20 questions
22. Rationalising limit of root x plus a minus root a over x 22 questions
23. The sandwich theorem 30 questions
24. Applications of continuity cost functions 30 questions
25. Trigonometric limits at other points and mixed forms 30 questions
26. Finding constants to make a function continuous 30 questions
27. Limit of xⁿ minus aⁿ over x minus a 25 questions
28. Expressing sequence limits in terms of e 30 questions
29. Limit of a sequence convergent and divergent 22 questions
30. The limit of (aˣ − 1)/x and logarithms 30 questions
31. Exercise 12.1 algebraic limits and mixed items 24 questions
32. Meaning of x approaches zero, infinity and a 23 questions
33. Limits at infinity of rational functions 30 questions
34. Exponential limits at infinity and e to the power x 30 questions
35. Proof that sin θ over θ tends to one 30 questions
36. Criterion for the existence of a limit 30 questions
37. Limit of a sequence rational in n 24 questions

Sample questions: Limit and Continuity

10 sample questions with the correct answer and an explanation. Try to answer each one before reading the explanation.

Q1. Express lim(n → +∞) (1 + 1/(3n))⁴ⁿ in terms of e.

• A. exp(4/3)(correct answer)
• B. exp(8/3)
• C. exp(1/3)
• D. exp(7/3)

Answer: A.Take k = 1/3 and m = 4. With u = n/k the base is 1 + 1/u and the power 4n equals km·u = 4/3·u. So the expression is [(1 + 1/u)ᵘ] raised to the power 4/3; since (1 + 1/u)ᵘ → e (also for u → −∞), the limit is e to the power 4/3, i.e. exp(4/3).

Q2. Express lim(n → +∞) (1 + 1/(2n))³ⁿ in terms of e.

• A. exp(3/2)(correct answer)
• B. exp(−3/2)
• C. exp(1/2)
• D. exp(5/2)

Answer: A.Take k = 1/2 and m = 3. With u = n/k the base is 1 + 1/u and the power 3n equals km·u = 3/2·u. So the expression is [(1 + 1/u)ᵘ] raised to the power 3/2; since (1 + 1/u)ᵘ → e (also for u → −∞), the limit is e to the power 3/2, i.e. exp(3/2).

Q3. Express lim(n → +∞) (1 + 1/(4n))³ⁿ in terms of e.

• A. exp(−3/4)
• B. exp(1/4)
• C. exp(7/4)
• D. exp(3/4)(correct answer)

Answer: D.Take k = 1/4 and m = 3. With u = n/k the base is 1 + 1/u and the power 3n equals km·u = 3/4·u. So the expression is [(1 + 1/u)ᵘ] raised to the power 3/4; since (1 + 1/u)ᵘ → e (also for u → −∞), the limit is e to the power 3/4, i.e. exp(3/4).

Q4. Express lim(x → 0) (1 + 5x) raised to the power 1/2/x in terms of e.

• A. exp(5/2)(correct answer)
• B. exp(7/2)
• C. exp(11/2)
• D. exp(1/2)

Answer: A.Take k = 5 and m = 1/2. With u = kx → 0 the base is 1 + u and the power 1/2/x equals km/u = 5/2/u. The expression is [(1 + u) raised to 1/u] raised to the power 5/2, and (1 + u) raised to 1/u → e, so the limit is e to the power 5/2, i.e. exp(5/2).

Q5. Express lim(x → 0) (1 − x) raised to the power 1/2/x in terms of e.

• A. exp(−1/2)(correct answer)
• B. e⁻¹
• C. 1
• D. exp(1/2)

Answer: A.Take k = −1 and m = 1/2. With u = kx → 0 the base is 1 + u and the power 1/2/x equals km/u = −1/2/u. The expression is [(1 + u) raised to 1/u] raised to the power −1/2, and (1 + u) raised to 1/u → e, so the limit is e to the power −1/2, i.e. exp(−1/2).

Q6. Express lim(x → 0) (1 + x/2) raised to the power 3/x in terms of e.

• A. exp(1/2)
• B. exp(3/2)(correct answer)
• C. exp(7/2)
• D. exp(5/2)

Answer: B.Take k = 1/2 and m = 3. With u = kx → 0 the base is 1 + u and the power 3/x equals km/u = 3/2/u. The expression is [(1 + u) raised to 1/u] raised to the power 3/2, and (1 + u) raised to 1/u → e, so the limit is e to the power 3/2, i.e. exp(3/2).

Q7. Express lim(x → 0) (1 + x) raised to the power 1/2/x in terms of e.

• A. exp(3/2)
• B. 1
• C. exp(−1/2)
• D. exp(1/2)(correct answer)

Answer: D.Take k = 1 and m = 1/2. With u = kx → 0 the base is 1 + u and the power 1/2/x equals km/u = 1/2/u. The expression is [(1 + u) raised to 1/u] raised to the power 1/2, and (1 + u) raised to 1/u → e, so the limit is e to the power 1/2, i.e. exp(1/2).

Q8. Express lim(x → 0) (1 + x/3) raised to the power 1/2/x in terms of e.

• A. exp(1/6)(correct answer)
• B. exp(7/6)
• C. exp(3/2)
• D. exp(5/6)

Answer: A.Take k = 1/3 and m = 1/2. With u = kx → 0 the base is 1 + u and the power 1/2/x equals km/u = 1/6/u. The expression is [(1 + u) raised to 1/u] raised to the power 1/6, and (1 + u) raised to 1/u → e, so the limit is e to the power 1/6, i.e. exp(1/6).

Q9. Express lim(x → 0) (1 + x/2) raised to the power 5/x in terms of e.

• A. exp(7/2)
• B. exp(5/2)(correct answer)
• C. exp(−5/2)
• D. exp(1/2)

Answer: B.Take k = 1/2 and m = 5. With u = kx → 0 the base is 1 + u and the power 5/x equals km/u = 5/2/u. The expression is [(1 + u) raised to 1/u] raised to the power 5/2, and (1 + u) raised to 1/u → e, so the limit is e to the power 5/2, i.e. exp(5/2).

Q10. Express lim(x → 0) (1 + x/3) raised to the power 2/x in terms of e.

• A. exp(7/3)
• B. exp(5/3)
• C. exp(2/3)(correct answer)
• D. exp(4/3)

Answer: C.Take k = 1/3 and m = 2. With u = kx → 0 the base is 1 + u and the power 2/x equals km/u = 2/3/u. The expression is [(1 + u) raised to 1/u] raised to the power 2/3, and (1 + u) raised to 1/u → e, so the limit is e to the power 2/3, i.e. exp(2/3).