Question

Match List I with List II

LIST I	LIST II
(A) limx→1(1 − x)1/x	(I) e
(B) limx→0 1/x ln(1 − x)	(II) 1
(C) limx→0 (1 + x2)1/x	(III) 0
(D) limx→∞ (1 + 1/x)x	(IV) 2

• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
• (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
• (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
• (A) - (IV), (B) - (II), (C) - (I), (D) - (III)

Hint:

Use known limit properties and approximations, such as limn→∞(1 + k/n)^n = e^k, to simplify computations.

Answer 1

The Correct Option is D

(A) limn→∞(1 − 1/n)^2n = e−2, so (A) matches (IV). (B) limx→1(1 − x^2)[log(1 − x)]−1 = e, so (B) matches (II). (C) limx→0(1 + x^2)e−x = 1, so (C) matches (I). (D) limx→∞(1 + 2/x)^x = e^2, so (D) matches (III).