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

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**As you have not specified what x tends to, I am taking it to be 0.**

The value of lim x-->0 [(3x+2)/(2x+1)]^[(2x+1)/(x+4)] is required.

lim x-->0 [(3x+2)/(2x+1)]^[(2x+1)/(x+4)]

substitute x = 0,

=> [(3*0+2)/(2*0+1)]^[(2*0+1)/(0+4)]

=> [(2/1)]^[(1/4)]

=> 2^(1/4)

**The value of lim x-->0 [(3x+2)/(2x+1)]^[(2x+1)/(x+4)] = 2^(1/4)**

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**