Question-126056

Question Number 126056 by TITA last updated on 16/Dec/20

Commented by TITA last updated on 16/Dec/20

please help

Answered by Olaf last updated on 16/Dec/20

\frac{x^{x} - 1}{\ln (x + 1)} = \frac{e^{x \ln x} - 1}{\ln (x + 1)}

La limite en 0^{+} est de la forme \frac{0^{+}}{0^{+}} .

C^{'} est une forme indeterminee .

On utilise la regle de l^{'} hopital :

\lim_{x \to 0^{+}} \frac{f (x)}{g (x)} = \lim_{x \to 0^{+}} \frac{f^{'} (x)}{g^{'} (x)}

donc on cherche :

\lim_{x \to 0^{+}} \frac{(1 . \ln x + x . \frac{1}{x}) e^{x \ln x}}{1 / (x + 1)}

\lim_{x \to 0^{+}} (x + 1) (\ln x + 1) e^{x \ln x}

de la forme 1 \times (- \infty) \times 1 donc - \infty

Answered by Dwaipayan Shikari last updated on 16/Dec/20

\lim_{x \to 0^{+}} \frac{e^{x l o g x} - 1}{l o g (x + 1)} = \frac{1 + x l o g (x) - 1}{x} \lim_{x \to 0} l o g (x + 1) = x a n d e^{x} = x + 1

= l o g (x) \to - \infty

Answered by 676597498 last updated on 17/Dec/20

a s x \to 0

l n (x + 1) \to x - \frac{x^{2}}{2!} + ξ (x)

\lim_{x \to 0^{+}} \frac{x^{x} - 1}{l n (x + 1)} = \lim_{x \to 0^{+}} \frac{x^{x} - 1}{x (1 - \frac{x}{2})}

= \lim_{x \to 0^{+}} \frac{x^{x - 1} - \frac{1}{x}}{1 - \frac{x}{2}} = - \infty