Limit of f(x)= 1/x + 1/(e^x-1) as x approaches 0 and infinity

B

BJU

New member
Joined
Jul 15, 2016
Messages
21
While solving some other problems, I have encountered the function

f(x) = 1/x + 1/(e^x - 1)

From plotting the function, I can see that the function approaches 1/2 from below when x->+0 and 0 from above when x-> infinity. However, I have no idea how to prove this since both individual terms approach infinity when x->0 and 0 when x-> infinity.

Any suggestions about how to approach a proof would be very appreciated, thanks!

BJ
 
BJU said:
While solving some other problems, I have encountered the function

f(x) = 1/x + 1/(e^x - 1)

From plotting the function, I can see that the function approaches 1/2 from below when x->+0 and 0 from above when x-> infinity. However, I have no idea how to prove this since both individual terms approach infinity when x->0 and 0 when x-> infinity.

Any suggestions about how to approach a proof would be very appreciated, thanks!

BJ
Click to expand...

Have you studied L'Hospital's rule?
 
I have never heard of that rule, but I will have a look. Many thanks!
 
Ok, for anybody interested in how this works:

L'Hôpital's rule says if limit(x->a)=f(x)/g(x) = 0/0 or infty/infty then limit(x->a)=f(x)/g(x)=f'(x)/g'(x)

So, my case, first, turn the term into a single fraction:

1/x - 1/(e^x-1) = (e^x - 1 - x)/(x(e^x))

For limit(x->0) f'(x)/g'(x) = (e^x - 1)/(x e^x + e^x - 1)

For x=0, this gives (1 - 1)/(0 + 1 - 1) = 0/0

So apply L'Hôpital's rule again

For limit(x->0) f''(x)/g''(x) = e^x/(2 E^x + E^x x)

For x=0, this gives (1)/(2 + 0) = 1/2

So the function indeed goes to 1/2 as x approaches 0.

Yay!

Many thanks again to Subhotosh Khan for the quick and very helpful pointer!
 
Thank you very much for sharing your completed solution.
 
Wait a minute!

BJU said:
Ok, for anybody interested in how this works:

L'Hôpital's rule says if limit(x->a)=f(x)/g(x) = 0/0 or infty/infty then limit(x->a)=f(x)/g(x)=f'(x)/g'(x)

So, my case, first, turn the term into a single fraction:

1/x - 1/(e^x-1) = (e^x - 1 - x)/(x(e^x))

For limit(x->0) f'(x)/g'(x) = (e^x - 1)/(x e^x + e^x - 1)

For x=0, this gives (1 - 1)/(0 + 1 - 1) = 0/0

So apply L'Hôpital's rule again

For limit(x->0) f''(x)/g''(x) = e^x/(2 E^x + E^x x)

For x=0, this gives (1)/(2 + 0) = 1/2

So the function indeed goes to 1/2 as x approaches 0.

Yay!

Many thanks again to Subhotosh Khan for the quick and very helpful pointer!
Click to expand...
==========================
The original had 1/x PLUS 1/(e^x - 1). No wonder you had trouble.
 
You must log in or register to reply here.
Top