More practice questions

5. Q5. Evaluate the limit as x → 0 of (sin ax + bx)/(ax + sin bx), where a, b and a + b are not 0 (NCERT Exercise 12.1, Q20).

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   Divide top and bottom by x. The top becomes a × (sin ax)/(ax) + b → a + b, and the bottom a + b × (sin bx)/(bx) → a + b.

   So the limit is (a + b)/(a + b) = 1.

6. Q6. Evaluate lim_x→0 (sin 3x)/(5x).

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   (3/5) × (sin 3x)/(3x) → 3/5.

7. Q7. Evaluate lim_x→0 (1 − cos 3x)/x².

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   1 − cos 3x = 2 sin²(3x/2), so the fraction is (9/2) × [sin(3x/2)/(3x/2)]² → 9/2.

8. Q8. Evaluate lim_x→0 (tan 3x)/(sin 2x).

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   3 × [(tan 3x)/(3x)] ÷ (2 × [(sin 2x)/(2x)]) → 3/2.

9. Q9. Evaluate lim_x→π (sin x)/(π − x).

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   Put y = π − x → 0: sin x = sin(π − y) = sin y, so the limit is lim (sin y)/y = 1.

10. Q10. Use the sandwich theorem to find lim_x→0 x sin(1/x).

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    For x ≠ 0, |sin(1/x)| ≤ 1, so −|x| ≤ x sin(1/x) ≤ |x| (not −x ≤ … ≤ x, which fails for x < 0). Both bounds tend to 0, so the limit is 0, although sin(1/x) itself has no limit at 0.

11. Q11. Evaluate lim_x→0 (tan x − sin x)/x³.

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    tan x − sin x = tan x(1 − cos x), so the fraction is [(tan x)/x] × [(1 − cos x)/x²] → 1 × 1/2 = 1/2.

12. Q12. With x measured in degrees, find lim_x→0 (1 − cos x°)/x².

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    Put u = πx/180, so u → 0, cos x° = cos u and x = 180u/π. The fraction is

    (π/180)² × (1 − cos u)/u²

    → (π/180)² × 1/2 = π²/64 800.