Question

what is limit of `x*log((1+e^x)/((-1)+e^x)))` as `x` approaches infinity ??

Answer 1

`0`

Explanation:

`(e^x + 1)/(e^x - 1) = (1 + e^-x)/(1 - e^-x)`
`= (1 + e^-x)^2/((1 - e^-x)(1 + e^-x))`
`= (1 + e^-x)^2/(1 - e^(-2 x))`
`= (1 + 2 e^-x + e^(-2 x))/(1 - e^(- 2 x))`
`~~ 1 + 2 e^-x`

`log(1+x) = x - x^2/2 + x^3/3 - ..." (Taylor series of ln(1+x))"`

`=> log((e^x + 1)/(e^x - 1)) ~~ 2 e^-x - 2 e^(-2 x) + ...`

`=> x log( ... ) ~~ 2 x e^-x -> 0" for "x -> oo`

Answer 2

`lim_(x->oo)(x*log[(1+e^x)/(-1+e^x)])=0`

Explanation:

Let's think about it this way:

As `x` gets really, really, large, `e^x` will be so large that numbers like `1` and `-1` will barely matter.

We can, therefore, ignore it.

We are left with `e^x/e^x` Since we are dividing same numbers, we conclude that `lim_(x->oo)(1+e^x)/(-1+e^x)=1`

We have:

`x*log(1)` Simplify

`=>x*0`

`=>0` The answer is `0`.

Just note that this method may not work as you go to harder versions of limit.