More practice questions

5. Q5. Evaluate lim_x→3 (x⁴ − 81)/(2x² − 5x − 3) (NCERT Exercise 12.1, Q8).

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   x⁴ − 81 = (x − 3)(x + 3)(x² + 9) and 2x² − 5x − 3 = (x − 3)(2x + 1). After cancelling, 6 × 18/7 = 108/7.

6. Q6. Evaluate lim_x→0 (ax + b)/(cx + 1) (NCERT Exercise 12.1, Q9).

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   The denominator tends to 1, so the limit is b.

7. Q7. Evaluate the limit as x → 1 of (ax² + bx + c)/(cx² + bx + a), where a + b + c ≠ 0 (NCERT Exercise 12.1, Q11).

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   (a + b + c)/(c + b + a) = 1.

8. Q8. Evaluate lim_x→−2 (1/x + 1/2)/(x + 2) (NCERT Exercise 12.1, Q12).

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   1/x + 1/2 = (x + 2)/(2x), so the expression is 1/(2x) for x ≠ −2. The limit is −1/4.

9. Q9. f(x) = 2x + 3 for x ≤ 0 and 3(x + 1) for x > 0. Find lim_x→0 f(x) and lim_x→1 f(x) (NCERT Exercise 12.1, Q23).

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   At 0 both one-sided limits are 3, so the limit is 3. Near 1 only the second formula applies: the limit is 3 × 2 = 6.

10. Q10. f(x) = x² − 1 for x ≤ 1 and −x² − 1 for x > 1. Find lim_x→1 f(x) (NCERT Exercise 12.1, Q24).

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    The left-hand limit is 0 and the right-hand limit is −2, so the limit does not exist.

11. Q11. Evaluate lim_x→0 f(x) for f(x) = |x|/x, and for f(x) = x/|x|, each with f(0) = 0 (NCERT Exercise 12.1, Q25 and Q26).

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    Both functions are 1 for x > 0 and −1 for x < 0. The one-sided limits are 1 and −1, so neither limit exists.

12. Q12. Find lim_x→5 (|x| − 5) (NCERT Exercise 12.1, Q27).

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    Near 5, |x| = x, so the limit is 5 − 5 = 0.

13. Q13. f(x) = (x − a₁)(x − a₂)…(x − a_n). Find lim_x→a1 f(x), and lim_x→a f(x) for a ≠ a₁, …, a_n (NCERT Exercise 12.1, Q29).

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    f is a polynomial, so the limits are its values: 0 at a₁, and (a − a₁)(a − a₂)…(a − a_n) at a.

14. Q14. f(x) = |x| + 1 for x < 0, 0 for x = 0 and |x| − 1 for x > 0. For which a does lim_x→a f(x) exist (NCERT Exercise 12.1, Q30)?

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    At 0 the one-sided limits are 1 and −1, so there is no limit. At every a ≠ 0, f is given near a by one formula, so the limit exists. Answer: every a ≠ 0.

15. Q15. f(x) = mx² + n for x < 0, nx + m for 0 ≤ x ≤ 1 and nx³ + m for x > 1. For which integers m and n do lim_x→0 f(x) and lim_x→1 f(x) both exist (NCERT Exercise 12.1, Q32)?

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    At 0: left n, right m, so m = n is needed. At 1: both sides give n + m, always. So both limits exist exactly when m = n.

16. Q16. Evaluate lim_x→0 (√(x + 4) − 2)/x.

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    Multiply top and bottom by √(x + 4) + 2: the expression becomes 1/(√(x + 4) + 2) for x ≠ 0, which tends to 1/4. (Theorem 2 with n = 1/2 and a = 4 gives the same.)