# 1/(e^x-1) limit

1. Nov 7, 2013

### alejandrito29

Hello ,

i need find the value of $$1/(e^x-1)$$, when x is small. The answer is 1/x.

if i use taylor series $$1/(e^x-1) \approx 1/(1+x-1) =1/x$$

but

¿there is a way to obtain the same answer using $$\lim_{x \to 0}1/(e^x-1)$$ ???

2. Nov 7, 2013

### Staff: Mentor

That limit does not exist. You can evaluate $$\lim_{x \to 0} \frac{x}{e^x-1} = 1$$ and use this to get the approximation 1/x.

3. Nov 7, 2013

### alejandrito29

thanks

why the limit does not exist?,

why $$\lim_{x \to 0} \frac{x}{e^x-1} = 1$$???.

Sorry , but i don't understand

4. Nov 7, 2013

### Staff: Mentor

The numerator is constant, the denominator goes to zero. It does not exist in the same way as the limit of 1/x for x -> 0 does not exist.

Calculate the limit, and you'll see it.

5. Nov 7, 2013

### Staff: Mentor

Use the full Maclaurin expansion for ex - 1, and then factor x out of the denominator.

6. Nov 7, 2013

### alejandrito29

thanks, very thanks

and

¿there is a way to obtain an answer equal to x using

$$\lim_{x \to \infty} 1/ \Big (exp (1/x) -1 \Big )$$ ???

7. Nov 7, 2013

### Staff: Mentor

No. If the limit does not exist, it does not matter how you try to calculate it.