Limits of indeterminate forms $(-\infty)(0)$

Question

Can anybody give me a detailed solution on how $$\lim_{x \to \infty} -x\left(1-e^{-\frac{1}{1+x}}\right) = -1?$$

I understand that separately the limit of each factor is $(-\infty)(0)$ but I could not proceed to getting $-1$. I also tried L'hopitals but I only got as far as

$$\lim_{x \to \infty} \frac{1}{-(1-e^{-\frac{1}{1+x}})^{-2}(-e^{-\frac{1}{1+x}})(-\frac{1}{(1+x)^2})} = \frac{-1}{(0)(1)(0)}.$$

Any leads would be helpful. Thank you!

Answer 1

Using limit written in comment, we have

$$x\left(e^{-\frac{1}{1+x}}-1\right)=x\cdot\frac{\left(e^{-\frac{1}{1+x}}-1\right)}{-\frac{1}{1+x}}\cdot\left(-\frac{1}{1+x}\right) \sim x\cdot \frac{-1}{1+x}\to -1$$ for $x\to\infty$.

Answer 2

Let $u=\frac1{1+x}$. Then we have $$\lim_{u\to 0}\left(\frac1u-1\right)(e^{-u}-1) = \lim_{u\to 0}\frac{e^{-u}-1}u,$$ since $\lim\limits_{u\to 0} (e^{-u}-1)=0$. Now finish by recognizing the definition of the derivative.

Answer 3

$$\begin{array}{l} {\rm\text{Since we knew}}:\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1\\ \displaystyle\Rightarrow - \mathop {\lim }\limits_{x \to \infty } x\left( {1 - {e^{ - \frac{1}{{1 + x}}}}} \right) = - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{ - \frac{1}{{1 + x}}}} - 1}}{{ - \frac{1}{{1 + x}}}}} \right).\left( {\frac{x}{{1 + x}}} \right) = - \mathop {\lim }\limits_{x \to \infty } \frac{x}{{1 + x}}\\ = \displaystyle - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + \frac{1}{x}}} = - 1 \end{array}$$

Answer 4

Another way, using that as $t\to 0$ we have $ 1+t \le e^t\le 1+t+t^2$ we get

$$-\frac{x}{1+x} \le x\left(e^{-\frac{1}{1+x}}-1\right)\le -\frac{x}{1+x}+\frac{x}{(1+x)^2}$$

which by squeeze theorem leads to $x\left(e^{-\frac{1}{1+x}}-1\right)\to -1$.