How do you find the limit of #[1/e^(x +1)]^sqrtx # as x approaches 0?

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Answer 1 · 2016-08-14T21:44:29

#1#

#lim_(x to 0) [1/e^(x +1)]^sqrtx#

#=lim_(x to 0) exp ( ln [1/e^(x +1)]^sqrtx )#

and because #e^x# is continuous for all #x#
#=exp ( lim_(x to 0) ln [1/e^(x +1)]^sqrtx )#

re-arranging the log term
#=exp ( lim_(x to 0) sqrt x ln (e^(-(x +1))) )#

ditto
#=exp ( lim_(x to 0) sqrt x (-(x +1))) )#

limt of product is product of limits if functions continuous through limits
#=exp ( -lim_(x to 0) sqrt x * lim_(x to 0) (x + 1) )#

#=e^ (- 0 * 1 )#

# = 1#