Finding Limit of 1/(e^x-1) at x=0

[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 $$\lim_{x \to 0}1/(e^x-1)$$ ?

 

[mfb]

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.

 

[alejandrito29]

That limit does not exist.

thanks

why the limit does not exist?,

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.

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

Sorry , but i don't understand

 

[mfb]

why the limit does not exist?,

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.

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

Calculate the limit, and you'll see it.

 

[Mark44]

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

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

 

[alejandrito29]

thanks, very thanks

and

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

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

 

[mfb]

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