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

Answer:

# 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 #a^x rarr 0# as #x rarr oo# provided #|a| lt 1#, And so:

# L = (1-0+0)/(4-0-0) #

# \ \ = 1/4 #