Limits
The value of \lim_{x \to 0} \frac{\ln(1 + 2x)}{\sin 3x} is required; choose the correct evaluation using standard limit equivalences.
Select the correct option:
Solution
32
Near zero the standard equivalences \ln(1+u) \sim u and \sin v \sim v let us replace transcendental functions by their leading linear parts, the cleanest tool for such 0/0 quotients. With u = 2x and v = 3x, the numerator behaves like 2x and the denominator like 3x, so the ratio tends to 2x/3x = 2/3. More formally, rewrite as \frac{\ln(1+2x)}{2x} \cdot \frac{3x}{\sin 3x} \cdot \frac{2x}{3x}, where the first factor tends to 1, the second to 1, and the third equals 2/3. Hence the limit is 2/3. Option 0 wrongly assumes both parts vanish without comparing their rates. Option 3/2 inverts the ratio, dividing 3 by 2 instead of 2 by 3. Option 1 forgets the differing coefficients of x in the two arguments. The governing JEE pattern is the use of first-order equivalences for logarithmic and sine limits. Plausibility check: at x = 0.001 the numerator \approx 0.0019998 and the denominator \approx 0.0029999, whose ratio \approx 0.6667 matches 2/3 to four places, confirming the analytic result.
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- limits
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
32
Near zero the standard equivalences \ln(1+u) \sim u and \sin v \sim v let us replace transcendental functions by their leading linear parts, the cleanest tool for such 0/0 quotients. With u = 2x and v = 3x, the numerator behaves like 2x and the denominator like 3x, so the ratio tends to 2x/3x = 2/3. More formally, rewrite as \frac{\ln(1+2x)}{2x} \cdot \frac{3x}{\sin 3x} \cdot \frac{2x}{3x}, where the first factor tends to 1, the second to 1, and the third equals 2/3. Hence the limit is 2/3. Option 0 wrongly assumes both parts vanish without comparing their rates. Option 3/2 inverts the ratio, dividing 3 by 2 instead of 2 by 3. Option 1 forgets the differing coefficients of x in the two arguments. The governing JEE pattern is the use of first-order equivalences for logarithmic and sine limits. Plausibility check: at x = 0.001 the numerator \approx 0.0019998 and the denominator \approx 0.0029999, whose ratio \approx 0.6667 matches 2/3 to four places, confirming the analytic result.
This easy difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of limits. It appeared in the 2025 exam.
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