سوال ۲۷ کنکور ارشد علوم کامپیوتر ۱۴۰۳

This is a $\tfrac00$ form; Taylor-expand to the leading power of $t$.

Denominator. $\ln(1+t) - t = \left(t - \tfrac{t^2}{2} + \cdots\right) - t = -\tfrac{t^2}{2} + O(t^3).$

Numerator. Write $(1+t)^{1+t} = e^{(1+t)\ln(1+t)}$. Since $(1+t)\ln(1+t) = t + \tfrac{t^2}{2} - \tfrac{t^3}{6} + \cdots,$ exponentiating gives $(1+t)^{1+t} = 1 + t + t^2 + \tfrac{t^3}{2} + \cdots$. Hence $(1+t)^{1+t} - t - 1 = t^2 + O(t^3).$

Ratio. $\dfrac{t^2 + \cdots}{-\tfrac{t^2}{2} + \cdots} \to \dfrac{1}{-1/2} = -2.$

The limit is $-2$ (option 4).

Exam trick: repeated L'Hôpital here is brutal. Two Taylor expansions to order $t^2$ settle it instantly.