Q.63 The value of the limit lim_x→∞ (x/2) ln (1 + 2024/x) is ______.

The Limit lim_x→∞ (x/2) ln(1 + 2024/x) Equals 1012

Limit Evaluation

Direct substitution yields ∞ ⋅ 0, an indeterminate form. Rewrite as:

½ lim_x→∞ x ln(1 + 2024/x)

or equivalently:

½ lim_x→∞ ln(1 + 2024/x) / (1/x), now a 0/0 form.

Apply L’Hôpital’s Rule

Differentiate numerator:

d/dx [ln(1+2024/x)] = −2024/(x² + 2024x)

Differentiate denominator:

d/dx [1/x] = −1/x²

Thus the limit becomes:

½ lim_x→∞ (−2024/(x²+2024x)) / (−1/x²)

= ½ lim_x→∞ 2024x²/(x²+2024x)

= ½ × 2024 = 1012

Alternative Standard Limit Method

Use lim_u→0 ln(1+u)/u = 1 with u = 2024/x → 0:

½ × 2024 = 1012

Introduction

Competitive exams like IIT JAM test indeterminate limits such as:

lim_x→∞ (x/2) ln(1 + 2024/x)

This ∞·0 form resolves cleanly to 1012 using standard calculus tools such as L’Hôpital’s rule and the limit definition of logarithms.

Alternative Methods

Taylor Expansion

ln(1+u) ≈ u as u → 0 with u = 2024/x:

(x/2)·(2024/x) = 1012

Substitution Method

Let t = 1/x → 0:

lim_t→0+ ½ · ln(1 + 2024t)/t = ½ × 2024 = 1012

Exam Note

Answer is a fill-in-the-blank: 1012.

Common errors include forgetting the 1/2 factor or misapplying substitution.