How do you find the limit of #((16x)/(16x+3))^(4x)# as x approaches infinity?

2 Answers
May 17, 2016

The limit is #e^(-3/4)#.

Explanation:

#((16x)/(16x+3))^(4x) #,

#lim_(xrarroo)((16x)/(16x+3))^(4x)# has indeterminate form #1^oo#.

We'll try to find the limit of the #ln# of the expression and use continuity of the exponential function to get

#lim_(xrarroo)((16x)/(16x+3))^(4x) = e^(lim_(xrarroo)ln(((16x)/(16x+3)^)^(4x))#

#ln(((16x)/(16x+3))^(4x)) = 4xln((16x)/(16x+3))#

# = 4ln((16x)/(16x+3))/(1/x)#

As #xrarroo#, this has indeterminate form #0/0#, so we can use l"Hospital's rule.

#lim_(xrarroo)4ln((16x)/(16x+3))/(1/x) = 4lim_(xrarroo)((16x+3)/(16x)((16(16x+3)-(16x)(16))/(16x+3)^2))/(-1/x^2)#

# = 4lim_(xrarroo)(48/(16x(16x+3)))/(-1/x^2)#

# = 4lim_(xrarroo)(-48x^2)/(16x(16x+3)))#

# = 4((-48)/16^2) = 4((-3)/16) = -3/4#

So

#lim_(xrarroo)((16x)/(16x+3))^(4x) = e^(-3/4)#

May 17, 2016

This limit can be evaluated without resorting to l'Hospital's rule using #lim_(urarroo)(1+1/u)^u = e#

Explanation:

#((16x)/(16x+3))^(4x) = (((16x+3)-3)/(16x+3))^(4x)#

# = (1+(-3)/(16x+3))^(4x)#

# = (1+1/((-(16x+3))/3))^(4x)#

# = (1+1/((-(16x+3))/3))^(4x)#

We need to make the exponent equal to #(-(16x+3))/3# so we'll multiply the exponent by #(-(16x+3))/(12x)# and also by the reciprocal.

# = ((1+1/((-(16x+3))/3))^((4x)(-(16x+3))/(12x)))^((-12x)/(16x+3))#

# = ((1+1/((-(16x+3))/3))^((-(16x+3))/3))^((-12x)/(16x+3))#

As #xrarroo#, we have #((-(16x+3))/3)rarroo# and #(-12x)/(16x+3) rarr -3/4#

Consequently,

#lim_(xrarroo)((16x)/(16x+3))^(4x) = lim_(xrarroo)[(1+1/((-(16x+3))/3))^((-(16x+3))/3)]^((-12x)/(16x+3))#

# = [e]^(-3/4)#