Why is the limit of $\frac{1}{1-e^x}$ when $x$ approaches $0^+$ equal to $-\infty$? [closed]

$$\lim_{x\to0^+}\frac{1}{1-e^x} = -\infty$$

Why is the limit equal to $-\infty$ and not $\infty$? Isn't $e^x$ at $x=0$ equal to $1$? Thus, $\frac{1}{1-1}=\infty$.

Solution 1:

If $x>0$, then $e^x>1$, and therefore $1-e^x<0$. So, since $\lim_{x\to0^+}1-e^x=0$,$$\lim_{x\to0^+}\frac1{1-e^x}=-\infty.$$

Solution 2:

The ${+}$ part of ${\lim_{x\to 0^+}}$ means that $x$ is getting smaller, but it is still always positive. For example, you could plug in some numbers, $$ \frac{1}{1-e^{0.1}}\approx \frac{1}{-0.105}\approx -9.51 $$ $$ \frac{1}{1-e^{0.01}}\approx \frac{1}{-0.01005}\approx -99.5 $$ and so on. As you pointed out, ${e^x}$ is approaching $1$, but ${e^x > 1}$ for any positive $x$, and thus ${1 - e^x < 0}$ for any positive $x$. It then follows ${\frac{1}{1-e^x}}$ has to be negative for any positive $x$.