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

Question

Isabella Ackerman · Accepted Answer

#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#

Samantha Adkison · Answer

undefined To find the limit of [1/e^(x +1)]^sqrtx as x approaches 0, we can use the properties of limits and exponential functions.

First, let's simplify the expression.

[1/e^(x +1)]^sqrtx = e^(-sqrtx(x + 1))

Now, as x approaches 0, the term sqrtx approaches 0 as well.

Using the limit properties, we can rewrite the expression as:

e^(-sqrtx(x + 1)) = e^(-0(x + 1)) = e^0 = 1

Therefore, the limit of [1/e^(x +1)]^sqrtx as x approaches 0 is 1.