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Limit with xeˣ and log(1+x)

Short answer

By applying L’Hospital’s rule twice, the numerator’s second derivative at 0 is 1 and denominator’s is 2, giving limit 1/2. Series gives difference x²/2 + …

Evaluate lim_{x→0} (xeˣ − (1+x)log(1+x))/x².

Let L = lim_{x→0} [x eˣ − (1 + x) log(1 + x)]/x². Direct substitution gives 0/0. Apply L'Hospital's rule.

Derivative of the numerator: eˣ + x eˣ − log(1 + x) − 1. At x = 0 this is 1 + 0 − 0 − 1 = 0, and the denominator derivative is 2x → 0, so again 0/0.

second derivative of numerator:
  d/dx [eˣ + x eˣ − log(1+x) − 1] = 2eˣ + x eˣ − 1/(1+x)
second derivative of denominator: 2

At x = 0: 2 + 0 − 1 = 1, so L = 1/2.

(Series check: x eˣ = x + x² + ...; (1+x) log(1+x) = x + x²/2 + ...; difference = x²/2 + ...)

Answer: 1/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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