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

Question

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

Emily Ammerman · Accepted Answer

#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#

Samantha Adkison · Answer

#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#

Ezekiel Alexander · Answer

undefined To find the limit as x approaches infinity, we can analyze the exponents of the terms involved. In the numerator, the exponent of 3 is (x+5), while the exponent of 2 is (2x+1). In the denominator, the exponent of 3 is (x+1), and the exponent of 2 is (2x+4).

As x approaches infinity, the terms with smaller exponents become negligible compared to those with larger exponents. Therefore, we can ignore the terms involving 2 in both the numerator and denominator.

This simplifies the expression to (3^(x+5))/(3^(x+1)).

Using the properties of exponents, we can rewrite this as 3^(x+5-x-1), which simplifies to 3^4.

Thus, the limit of the given expression as x approaches infinity is 81.

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

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