Question

Limit of f(x) as x-> infinity, f(x) = x/(e^x-1)

Answer 1

So this limit appears to be giving an indeterminate form of: \(\Large \cfrac{\infty}{\infty}\)

Hmm, have you learned about `L'Hopital's Rule` yet?

Answer 2

L'Hopital's rule will simplify the problem, but it might be considered extra machinery to get an answer that doesn't require it.

It could be simpler to simply multiply the original equation by 1, like so:
\[f(x) = \frac{\frac{1}{e^x}}{ \frac{ 1 }{ e^x } } \times \frac{x}{e^x -1}\]

then solve for the equation for x as it tends to infinity. It will give the same answer as using L'Hopital's rule.

Answer 3

In case you can't read those fractions, its 1/e^x.

Answer 4

I dont quite understand, if I do the multiply by one thing don't I end up getting a 0 in the denominator when I simplify?

Answer 5

You don't even need to use L'hopital's rule for the problem. Just use relative rates of change. e^x grows significantly faster in comparison to 'x'. So even though both functions go to positive infinity as goes to infinity, one approaches it much faster than the other which ultimately results in the denominator being much higher than the numerator which effectively leads to a limit of 0 as x tends to positive infinity.

@zzr0ck3r @halppls

Answer 6

its true e^x expands fastest to infinity compared to x and 1 which is a constant and so, we have a larger number at the denominator than at the numerator which can be assumed to be 0/infinity which is zero.