How do you find the limit of #(2+x)^(1/x)# as x approaches #0^+#?

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Answer 1 · 2016-10-13T13:54:23

Rewrite as an exponential.

#f(x) = (2+x)^(1/x) = e^(1/xln(2+x)#

#lim_(xrarro^+)(2+x)^(1/x) = lim_(xrarr0^+) e^(1/xln(2+x)#

# = = e^(lim_(xrarr0^+) (ln(2+x)/x)#

Note that #lim_(xrarr0^+) ln(2+x)/x = oo#.

Furthermore, as #trarroo#, we have #e^t rarroo#.

Therefore,

#lim_(xrarro^+)(2+x)^(1/x) =oo#