Question #0f6bd

In this problem, we will make repeated use of the fact that due to the continuity of `e^x`, we have `lim_(x->oo)e^f(x) = e^(lim_(x->oo)f(x))`. We will also use the properties of logarithms that

• `e^(ln(x)) = x`
• `ln(a^x) = xln(a)`

along with L'Hopital's rule, the chain rule, and the derivatives `d/dxln(x) = 1/x` and `d/dxx^n = nx^(n-1)`

Proceeding...

`lim_(x->oo)e^((1+x^2)^(1/ln(x))) = e^[lim_(x->oo)(1+x^2)^(1/ln(x))]`

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

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

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

`lim_(x->oo)ln(1+x^2)/ln(x) = lim_(x->oo)(d/dxln(1+x^2))/(d/dxln(x))`

`=lim_(x->oo)((2x)/(1+x^2))/(1/x)`

`=lim_(x->oo)(2x^2)/(x^2+1)`

`=lim_(x->oo)2/(1+1/x^2)`

`=2`

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

`:.lim_(x->oo)e^((1+x^2)^(1/ln(x))) = e^(e^2)`

First making `y = x^2` we have

`(1+x^2)^(1/lnx)equiv(1+y)^(2/lny)`

calling now `u=lny`

`(1+y)^(2/lny)equiv(1+e^u)^(2/u) = (e^(-u)+1)e^2`

but `x->oo` implies in `y->oo` and
`y->oo` implies in `u->oo` so

`lim_(x->oo)(1+x^2)^(1/lnx)equivlim_(u->oo) (e^(-u)+1)e^2=e^2`

Finally

`lim_(x->oo)e^((1+x^2)^(1/lnx)) = e^(e^2)`