What is the limit of ${(1 + e^{x})}^{\frac{1}{x}}$ $(1+e^x)^(1/x)$ as x approaches infinity?

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Answer 1 · 2016-11-29T02:52:31

$lim_(x->oo)(1+e^x)^(1/x) = e$

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

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

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

The above step is valid due to the continuity of $f(x)=e^x$

Evaluating the limit in the exponent, we have

$lim_(x->oo)ln(1+e^x)/x$

$=lim_(x->oo)(d/dxln(1+e^x))/(d/dxx)$

The above step follows from apply L'Hopital's rule to a $oo/oo$ indeterminate form

$=lim_(x->oo)(e^x/(1+e^x))/1$

$=lim_(x->oo)e^x/(1+e^x)$

$=lim_(x->oo)1/(1+1/e^x)$

$=1/(1+1/oo)$

$=1/(1+0)$

$=1$

Substituting this back into the exponent, we get our final result:

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

$=e^1$

$=e$

What is the limit of (1+e^x)^(1/x)(1+ex)1x(1+e^x)^(1/x) as x approaches infinity?