How do you evaluate the limit #root3((x-3)/(5-x))# as x approaches #-oo#?

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Answer 1 · 2017-10-18T11:54:16

#-1# for #RR#

Writing the expression in the radical as:

#(x-3)/(-x+5)#

For limits to infinity we can disregard the constants:

#x/-x= -1#

#lim_(x-> -oo)root(3)(x-3)/(5-x)= root(3)(-1)#

Using De'Moivre's formula:

#-cos(2pi)+isin(2pi)= -1#

#-cos((2pi)/3)+isin((2pi)/3)= 1/2+1/2 i sqrt(3) #

#-cos((4pi)/3)+isin((4pi)/3) =1/2-1/2 i sqrt(3)#