How do you find the limit #lim (3^(x+5)-2^(2x+1))/(3^(x+1)-2^(2x+4))# as #x->oo#?

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Answer 1 · 2018-01-02T11:58:13

#1/8#

#(3^(x+5)-2^(2x+1))/(3^(x+1)-2^(2x+4))#

Rewriting:

#(3^(x+5)-4^(x+1/2))/(3^(x+1)-4^(x+2))#

Concentrating on the dominating terms of numerator and denominator:

#(-4^(x+1/2))/(-4^(x+2))=(4^(x+1/2))/(4^(x+2))#

Dividing:

#(4^(-3/2))=1/(4^(3/2))=1/8#

#lim_(x->oo)((3^(x+5)-2^(2x+1))/(3^(x+1)-2^(2x+4)))=1/8#

Answer 2 · 2018-01-02T16:12:17

#1/8#

#(3^(x+5)-2^(2x+1))/(3^(x+1)-2^(2x+4)) = (3^5 3^x - 2 * 4^x)/(3 * 3^x-2^4 4^x)#

# = (3^5 (3/4)^x - 2)/(3(3/4)^x - 16)#

As #xrarroo#, we have #(3/4)^xrarr0#, so the limit sought

is #(0-2)/(0-16) = 1/8#