Limite notável

[fff]

Boa tarde. Tenho dúvidas em calcular este limite.
\(\lim_{x \to -\infty} \frac{ln(e^{-x}-1)}{x}\)
Resposta: -1

[Man Utd]

\(\lim_{x \to -\infty} \frac{ln(e^{-x}-1)}{x}\)


\(u=e^{-x}-1 \;\; x=-ln(u+1) \;\; x \rightarrow -\infty \; , \; u \rightarrow +\infty\) :


\(-\lim_{u \to +\infty} \; \frac{\ln(u)}{\ln(u+1)}\)


\(-\lim_{u \to +\infty} \; \frac{\ln(u)}{\ln(u*(1+\frac{1}{u}))}\)


\(-\lim_{u \to +\infty} \; \frac{\ln(u)}{\ln(u)+\ln(1+\frac{1}{u}))}\)


\(-\left( \; \frac{\lim_{u \to +\infty} \; \ln(u)}{\lim_{u \to +\infty} \;\ln(u)+\lim_{u \to +\infty}\;\ln(1+\frac{1}{u}))} \right )\)


\(-\left( \; \frac{\lim_{u \to +\infty} \; \ln(u)}{\lim_{u \to +\infty} \;\ln(u)}\right )\)


\(-\left(\lim_{u \to +\infty} \; \frac{ \ln(u)}{\ln(u)}\right )\)


\(-(1)=\fbox{\fbox{-1}}\)