Q.56
For x₁ > 0 and x₂ > 0, the value of
lim_{x₁ → x₂} (x₁ − x₂) / [x₂ ln(x₁ / x₂)] is ___________.

Problem Statement

Evaluate the limit:

lim_{x1 → x2}
(x₂ ln(x₁/x₂)) / (x₁ − x₂)

with x₁ > 0 and x₂ > 0.

The result of this limit is 1/x₂.

Why It’s Indeterminate

Direct substitution gives

• Numerator: x₂ ln(1) = 0
• Denominator: x₁ − x₂ = 0

This forms 0/0, requiring further evaluation.

Step-by-Step Solution

1. Substitution Method

Let h = x₁ − x₂. As x₁ → x₂, h → 0.

x₁ = x₂ + h

x₁/x₂ = 1 + h/x₂

Limit becomes:

lim_h→0 x₂ ln(1 + h/x₂) / h

Rewrite:

= (1/x₂) lim_h→0 ln(1 + h/x₂) / (h/x₂)

Let u = h/x₂. Then

= (1/x₂) lim_u→0 ln(1 + u)/u

Using standard limit lim_u→0 ln(1+u)/u = 1:

= 1/x₂

2. Using L’Hôpital’s Rule

lim_x1→x2 x₂ ln(x₁/x₂) / (x₁ − x₂)

Differentiate numerator & denominator wrt x₁:

• d/dx₁[x₂ ln(x₁/x₂)] = x₂ · (1/x₁)
• d/dx₁[x₁ − x₂] = 1

Thus:

= lim_x1→x2 x₂/x₁ = 1/x₂

3. Series Expansion

ln(1+u) = u − u²/2 + O(u³)

ln(1+u)/u ≈ 1 − u/2 → 1 as u→0

Thus limit = 1/x₂.

Final Answer

lim_x1→x2 x₂ ln(x₁/x₂) / (x₁ − x₂) = 1/x₂

Why It Appears in Exams

Common in JEE, GATE, and calculus tests as a concept check for log limits.