How do you find the limit #lim (3^x-2^x+1)/(4*3^x-2^x-1)# as #x->oo#?
1 Answer
Jul 19, 2017
# lim_(x rarr oo) (3^x-2^x+1)/(4*3^x-2^x-1) =1/4#
Explanation:
We seek:
# L = lim_(x rarr oo) (3^x-2^x+1)/(4*3^x-2^x-1) #
We can manipulate the limits as follows:
# L = lim_(x rarr oo) (3^x-2^x+1)/(4*3^x-2^x-1) * (1/3^x)/(1/3^x) #
# \ \ = lim_(x rarr oo) ((1/3^x)(3^x-2^x+1))/((1/3^x)(4*3^x-2^x-1)) #
# \ \ = lim_(x rarr oo) (3^x/3^x-2^x/3^x+1/3^x)/(4*3^x/3^x-2^x/3^x-1/3^x) #
# \ \ = lim_(x rarr oo) (1-(2/3)^x+(1/3)^x)/(4*1-(2/3)^x-(1/3)^x) #
Now
# L = (1-0+0)/(4-0-0) #
# \ \ = 1/4 #