Question

Question #7c813

Answer 1

`lim_(xrarroo)((2x+3)/(2x+1))^(x+1)=e`

Explanation:

`L=lim_(xrarroo)((2x+3)/(2x+1))^(x+1)`

When we have a function with an exponent that's another function, it's best to take the natural logarithm of both sides:

`ln(L)=ln(lim_(xrarroo)((2x+3)/(2x+1))^(x+1))`

The natural logarithm can be moved inside the limit:

`ln(L)=lim_(xrarroo)ln(((2x+3)/(2x+1))^(x+1))`

Rewrite the function using `log(A^B)=Blog(A)`:

`ln(L)=lim_(xrarroo)(x+1)ln((2x+3)/(2x+1))`

This can be rewritten as a quotient:

`ln(L)=lim_(xrarroo)ln((2x+3)/(2x+1))/(1/(x+1))`

Note that `lim_(xrarroo)ln((2x+3)/(2x+1))=ln(2/2)=ln(1)=0` and `lim_(xrarroo)(1/(x+1))=0`, so this entire limit is in the indeterminate form `0/0`. As such, we can apply L'Hôpital's rule by taking the derivatives of the numerator and denominator:

`ln(L)=lim_(xrarroo)(d/dxln((2x+3)/(2x+1)))/(d/dx(1/(x+1)))`

Finding both of these limits individually:

`d/dxln((2x+3)/(2x+1))=d/dxln(2x+3)-d/dxln(2x+1)`

`color(white)(d/dxln((2x+3)/(2x+1)))=2/(2x+3)-2/(2x+1)`

`color(white)(d/dxln((2x+3)/(2x+1)))=(2(2x+1)-2(2x+3))/((2x+3)(2x+1))`

`color(white)(d/dxln((2x+3)/(2x+1)))=(-4)/((2x+3)(2x+1))`

`" "`

`d/dx(1/(x+1))=d/dx(x+1)^-1`

`color(white)(d/dx(1/(x+1)))=-(x+1)^-2d/dx(x+1)`

`color(white)(d/dx(1/(x+1)))=-(x+1)^-2`

`color(white)(d/dx(1/(x+1)))=-1/(x+1)^2`

Then:

`ln(L)=lim_(xrarroo)((-4)/((2x+3)(2x+1)))/(-1/(x+1)^2)`

Rewriting:

`ln(L)=lim_(xrarroo)(4(x+1)^2)/((2x+3)(2x+1))`

Both the numerator and denominator have a degree of `2` and leading coefficients of `4`, so:

`ln(L)=1`

Thus:

`L=e^1=e`

Answer 2

`e`

Explanation:

`((2 x + 3)/(2 x + 1))^(x + 1)=(2x)^(x+1)/(2x)^(x+1)((1+3/(2x))/(1+1/(2x)))^(x+1) =`
`=((1+3/(2x))/(1+1/(2x)))^(x+1) =(1+3/(2x))^(x+1)/(1+1/(2x))^(x+1)`

Now doing `z=(2x)/3` and `y=2x` we have

`((2 x + 3)/(2 x + 1))^(x + 1) = (1+1/z)^((3z)/2+1)/(1+1/y)^(y/2+1)=(1+1/z)^((3z)/2)/(1+1/y)^(y/2)((1+3/(2x))/(1+1/(2x)))` so

`((2 x + 3)/(2 x + 1))^(x + 1)=((1+1/z)^z)^(3/2)/((1+1/y)^y)^(1/2)((1+3/(2x))/(1+1/(2x)))`

knowing that `x->oo => {y ->oo, z->oo}` we have

`lim_(x->oo)((2 x + 3)/(2 x + 1))^(x + 1)=(lim_(z->oo)(1+1/z)^z)^(3/2)/(lim_(y->oo)(1+1/y)^y)^(1/2)lim_(x->oo)((1+3/(2x))/(1+1/(2x))) = e^(3/2)/e^(1/2) cdot 1 = e`