Jack ✨

*sonríe un poco* igualmente *lo mira directo* entonces... verdad o reto?

*se sienta en el sofá, extendiendo los brazos en el respaldo* reto. *sonríe de lado* Pero aviso que no me asusto fácil

*se sienta cerca de él* mm... te reto a que me digas algo que nunca le hayas dicho a nadie

*su mandíbula se tensa al verte reír con el otro*

*siento tu mirada pero sigo con mi amigo*

To find the limit as \( x \) approaches 1 from the positive side of \( (x-1)^x \), we start by noting that as \( x \) approaches 1 from the right, \( x - 1 \) approaches 0. This means we are dealing with the form \( 0^1 \), which is straightforward but can be verified using logarithms. First, let \( L = \lim_{x \to 1^+} (x - 1)^x \). Taking the natural logarithm of both sides, we get: \[ \ln L = \lim_{x \to 1^+} x \ln(x - 1) \] As \( x \) approaches 1 from the right, \( \ln(x - 1) \) approaches \( -\infty \). Therefore, the expression \( x \ln(x - 1) \) approaches \( 1 \cdot (-\infty) = -\infty \). Thus, \( \ln L = -\infty \), which implies \( L = e^{-\infty} = 0 \). To confirm, we substitute \( x = 1 + h \) where \( h \) approaches 0 from the positive side. The expression becomes: \[ \lim_{h \to 0^+} (h)^{1 + h} \] Taking the natural logarithm, we get: \[ \lim_{h \to 0^+} (1 + h) \ln h \] As \( h \to 0^+ \), \( \ln h \to -\infty \), so the expression approaches \( 1 \cdot (-\infty) = -\infty \). Exponentiating gives \( e^{-\infty} = 0 \). Another method involves rewriting the limit using \( t = x - 1 \), so as \( t \to 0^+ \): \[ \lim_{t \to 0^+} t^{t + 1} = \lim_{t \to 0^+} t \cdot t^t \] Since \( \lim_{t \to 0^+} t^t = 1 \), the limit simplifies to: \[ \lim_{t \to 0^+} t \cdot 1 = 0 \] Thus, the limit is confirmed to be 0 through multiple methods and numerical verification. \[ \boxed{0} \]

*te mira un poco confuso* Yo... soy Jack... y tu?

y tú quién eres

Ya te lo dije, soy Jack...

*se acerca un poco mas, y te susurra* te podrias sentar en mi regazo...?