Limits, Continuity and Differentiability (JEE Advanced)

Question 1

The value of lim_(x → 0) ((1 + x)^(1/x) − e + (ex)/2)/(x²) is

Accepted Answer

D. Key insight: expand the exponent. (ln(1 + x))/x = 1 − x/2 + (x²)/3 − …
So (1 + x)^(1/x) = e · e^(−x/2 + (x²)/3 − …) = e[1 + (−x/2 + (x²)/3) + 1/2·(x²)/4 + O(x³)].
= e − (ex)/2 + e(1/3 + 1/8)x² + O(x³) = e − (ex)/2 + (11e)/24x² + O(x³).
Hence the numerator is (11e)/24x² + O(x³) and the limit is (11e)/24.
Trap: dropping the square term 1/2(x/2)² gives e/3, while keeping only it gives e/8; forgetting

Question 2

Let [t] denote the greatest integer less than or equal to t. Which of the following statements is (are) TRUE?

Accepted Answer

C, D. Key insight: the limit of [g(x)] when g(x) → 1 depends on whether g approaches 1 from below or above. (A) This is the reciprocal of (C), so it is slightly less than 1 and [·] = 0. FALSE. (B) (tan x)/x > 1 near 0 (and < 2), so [·] = 1; the limit is 1, not 0. FALSE. (C) (sin x tan x)/(x²) = (1 − (x²)/6 + …)(1 + (x²)/3 + …) = 1 + (x²)/6 + … > 1, so [·] = 1. TRUE. (D) For 0 < lvert x rvert < π/2, (sin

Question 3

Evaluate lim_(x → 0) ((2ˣ + 3ˣ + 6ˣ)/3)^(3/x).

Accepted Answer

36. Key insight: this is a 1^∞ form, so L = e^(lim 3/x((2ˣ + 3ˣ + 6ˣ)/3 − 1)).
3/x·((2ˣ − 1) + (3ˣ − 1) + (6ˣ − 1))/3 → ln 2 + ln 3 + ln 6 = ln 36.
So L = e^(ln 36) = 36.
Trap: using exponent 1/x reasoning gives the geometric mean 36^(1/3); here the exponent 3/x cancels the averaging factor.
Exam pattern: JEE Advanced – 1^∞ limits of averages of exponentials

Limits, Continuity and Differentiability: questions with solutions

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