Math / CalculusQuestionlimx→−∞−x2+exx−1−ln(e−x+1)\lim _{x \rightarrow-\infty} \frac{-x^{2}+e x}{x-1}-\ln \left(e^{-x}+1\right)limx→−∞x−1−x2+ex−ln(e−x+1)Studdy Solutionإيجاد النهاية للتعبير المبسط:limx→−∞(e−1)xx−1=limx→−∞(e−1)1−1x=e−1\lim_{x \rightarrow -\infty} \frac{(e-1)x}{x-1} = \lim_{x \rightarrow -\infty} \frac{(e-1)}{1-\frac{1}{x}} = e-1limx→−∞x−1(e−1)x=limx→−∞1−x1(e−1)=e−1 النهاية النهائية هي:e−1\boxed{e-1}e−1View Full Solution - FreeWas this helpful?
