Question

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

Answer 1

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)`

Answer 2

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)`