Find the limit of the function [(3x+2)/(2x+1)]^[(2x+1)/(x+4)]; x approaches to infinite.

Since the variable x is present at base, also at exponent,
we'll evaluate the limit of base and the limit of
exponent:

lim [(3x+2)/(2x+1)]^lim
[(2x+1)/(x+4)]

We'll start with the limit of base and we'll
factorize by x both numerator and denominator.

lim
[(3x+2)/(2x+1)] = lim x*(3 + 2/x)/x*(2 + 1/x)

We'll
simplify and we'll get:

lim (3 + 2/x)/(2 + 1/x) = [lim 3 +
lim (2/x)]/[lim 2 + lim (1/x)]

[lim 3 + lim (2/x)]/[lim 2 +
lim (1/x)] = (3 + 2/infinite)/(2 + 1/infinite) = (3+0)/(2+0) =
3/2

We'll evaluate the limit of
exponent:

lim [(2x+1)/(x+4)] = lim x*(2 + 1/x)/x*(1 +
4/x)

lim x*(2 + 1/x)/x*(1 + 4/x) = lim (2 + 1/x)/(1 +
4/x)

lim (2 + 1/x)/(1 + 4/x) = [lim 2 + lim (1/x)]/[lim 1 +
lim (4/x)]

lim (2 + 1/x)/(1 + 4/x) =
2

The limit of the given function
is:

lim [(3x+2)/(2x+1)]^lim [(2x+1)/(x+4)] =
(3/2)^2

lim [(3x+2)/(2x+1)]^lim
[(2x+1)/(x+4)] = 9/4