Hello!

I am not sure why I am getting this problem wrong, and I keep getting six. Could someone be kind enough to show me how it's done?
Thank you very much in advance!

Find the limit. Use l'Hospital's Rule if appropriate.
lim ((5x − 2)/(5x + 4))^(5x + 1)
x→∞

Maybe try this:

http://www.youtube.com/watch?v=8CuBoOaiBfA

To find the limit of the given expression, we can apply l'Hospital's Rule if it is appropriate. In this case, let's check if we have an indeterminate form of the type 0^0 or ∞^0.

First, let's simplify the expression by multiplying the numerator and denominator by 1/(5x):

lim ((5x - 2)/(5x + 4))^(5x + 1)
x→∞

= lim ((5x - 2)/(5x + 4)) * (1/(5x))^((5x + 1)/(5x))
x→∞

Now, let's analyze the two factors separately.

The first factor, ((5x - 2)/(5x + 4)), does not have an indeterminate form as x approaches infinity. The limit of this factor can be found by substituting infinity into x:

lim ((5x - 2)/(5x + 4)) = (5∞ - 2)/(5∞ + 4) = (5∞)/(5∞) = 5/5 = 1
x→∞

The second factor, (1/(5x))^((5x + 1)/(5x)), is of the form 1^∞, which is an indeterminate form. So, we'll apply l'Hospital's Rule to this factor.

Taking the natural logarithm of the second factor, we get:

ln((1/(5x))^((5x + 1)/(5x))) = (5x + 1)/(5x) * ln(1/(5x))
= (5x + 1)/(5x) * (-ln(5x))
= -(5x + 1)/(5x) * ln(5x)

Now, let's determine the limit of this expression:

lim (-(5x + 1)/(5x) * ln(5x))
x→∞

Again, we have an indeterminate form (-∞ * ∞). So, we'll apply l'Hospital's Rule again.

Differentiating the numerator and denominator of the expression above, we get:

= -(5 * ln(5x) + 1 * (1/x))/(5)

Now, let's determine the limit of this expression:

lim (-(5 * ln(5x) + 1 * (1/x))/(5))
x→∞

This no longer has an indeterminate form and can be evaluated directly by substituting infinity for x:

= -(5 * ln(5∞) + 1 * (1/∞))/(5)
= -(5 * ln(∞) + 0)/(5)
= -(∞ + 0)/(5)
= -∞/5
= -∞

Now, we have evaluated the limit of the second factor.

Multiplying the limits of the two factors, we get:

lim ((5x - 2)/(5x + 4)) * (1/(5x))^((5x + 1)/(5x))
x→∞

= 1 * (-∞)
= -∞

Therefore, the limit of the given expression as x approaches infinity is -∞.