# What is lim x->infinity of x-ln(e^x+1)?

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### 1 Answer

Notice that if substitute oo for x in equation yields an inedetermination `oo-oo` , that is not accepted.

You need to factor out x such that:

`lim_(x-gtoo) x(1 - (ln(e^x+1))/x)`

You need to evaluate `lim_(x-gtoo) (ln(e^x+1))/x = oo/oo.`

You need to use l'Hospital's theorem such that:

`lim_(x-gtoo) (ln(e^x+1))/x = lim_(x-gtoo) ((ln(e^x+1))')/(x') `

`lim_(x-gtoo) (ln(e^x+1))/x = lim_(x-gtoo) (e^x)/(e^x+1) = oo/oo`

You need to use l'Hospital's theorem again such that:

`lim_(x-gtoo) (e^x)/(e^x+1) =lim_(x-gtoo) ((e^x)')/((e^x+1)') `

`lim_(x-gtoo) (e^x)/(e^x+1) = lim_(x-gtoo) (e^x)/(e^x) = 1`

`lim_(x-gtoo) x(1 - (ln(e^x+1))/x) = oo(1-1) = oo*0`

You should convert the indetermination `oo*0` into `0/0` to use l'Hospital's theorem such that:

`lim_(x-gtoo) (1 - (ln(e^x+1))/x) /(1/x) = lim_(x-gtoo) ( - (xe^x)/(e^x+1) + (ln(e^x+1)))/(-1/(x^2)) = oo`

**Hence, evaluating the limit to the given function yields**

` lim_(x-gtoo) (x-ln(e^x+1)) = oo.`

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