Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 62

Replacing oo for x in limit equation yields the nedetermination oo^o . You need to use the following technique, such that:
f(x) = (e^x + x)^(1/x)
You need to take logarithms both sides, such that:
ln f(x) = ln ((e^x + x)^(1/x))
Using the property of logarithms yields:
ln f(x) = (1/x)*ln ((e^x + x))
ln f(x) = (ln ((e^x + x)))/x
You need to evaluate the limit:
lim_(x->oo) ln f(x) = lim_(x->oo) (ln ((e^x + x)))/x = oo/oo
You need to use L'Hospital theorem:
lim_(x->oo) (ln ((e^x + x)))/x = lim_(x->oo) ((ln ((e^x + x)))')/(x')
lim_(x->oo) ((ln ((e^x + x)))')/(x')= lim_(x->oo) (((e^x + x)')/(e^x + x))/1
lim_(x->oo) ((e^x + x)')/(e^x + x) = lim_(x->oo) (e^x + 1)/(e^x + x)
You need to factor out e^x such that:
lim_(x->oo) (e^x + 1)/(e^x + x) = lim_(x->oo) (e^x(1 + 1/(e^x)))/(e^x(1 + x/(e^x)))
lim_(x->oo) (e^x(1 + 1/(e^x)))/(e^x(1 + x/(e^x))) = lim_(x->oo) (1 + 1/(e^x)))/((1 + x/(e^x)))
Since lim_(x->oo)1/(e^x) = 0 and lim_(x->oo) x/(e^x) = 0 yields:
lim_(x->oo) (1 + 1/(e^x)))/((1 + x/(e^x))) = 1
Hence, lim_(x->oo) ln f(x) = 1, such that lim_(x->oo) f(x) = e^1
Hence, evaluating the given limit, using l'Hospital rule and logarithm technique yields lim_(x->oo) (e^x + x)^(1/x) = e.