What is #lim_(x->oo) (1+x)^(1//x)#?

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Answer 1 · 2017-05-25T07:32:23

I get #1#, and also get it on Wolfram Alpha.

Well, let #u = 1/x# to see.

#=> L = lim_(x->oo) (1+x)^(1"/"x)#

#= lim_(u->0) (1 + 1/u)^(u)#

Since #lim_(u->0) 1/u# tends towards infinity, the #1# vanishes by then, so we might as well forget about that #1#.

#=> L = lim_(u->0) (1/u)^u#

This is of the form #oo^0#, and indeterminate. Let's try taking the #ln# of both sides, to see what we can do to that #u# exponent.

#ln L = ln lim_(u->0) (1/u)^u#

#= lim_(u->0) u ln (1/u)#

#= lim_(u->0) (ln (1/u))/(1"/"u)#

Since this is now of the form #oo/oo#, we can use L'Hopital's rule. Take the derivative of the numerator and denominator separately:

#ln L = lim_(u->0) (1/(1"/"u)cdotcancel(-1/u^2))/(cancel(-1/u^2))#

#= lim_(u->0) u = 0#

Therefore:

#color(blue)(L) = e^(lim_(u->0) u) = e^0#

#= color(blue)(1)#