Question #041b9

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Answer 1 · 2017-05-14T15:34:46

1

#2^(1/x)/(1+2^(1/x)) = 2^u/(1+2^u)# where #u# approaches to #+oo#

# 2^u/(1+2^u) = 1/(1/(2^u)+1)#

#u -> +oo <=> 2^u -> +oo#

then the limit will be #1/(0+1) = 1#

Answer 2 · 2017-05-14T15:42:41

#1.#

As #x to 0+, 1/x to +oo. #

Recall that, #"as "x to oo+, a^x to oo+," where "a gt 1.#

#:. 2^(1/x) to oo+," & so, "(1+2^(1/x)) to oo+," giving, "#

#lim_(x to 0+) 1/(1+2^(1/x))=0................(ast).#

#"Now, "2^(1/x)/(1+2^(1/x))={(1+2^(1/x))-1}/(1+2^(1/x))#

#=(1+2^(1/x))/(1+2^(1/x))-1/(1+2^(1/x))#

#=1-1/(1+2^(1/x)).#

#:.," by "(ast)," The Reqd. Lim.="1-0=1.#