` `

`=lim_(x->oo) (-8(2+1/x^2)^2)/((2-3/x)(2+5/x))=lim_(x->oo) (-8(2)^2)/(2*2) = -8`

This goes under the line

This we can solve by dividing both top and bottom by `x^2` to get

Forgot to change to math notation.

`lim_(x->oo) ((2x-3)/(2x+5))^(2x+1)`

We can take the ln of this using the log property. to get

`lim_(x->oo) (2x+1)ln((2x-3)/(2x+5))`

This limit is of the form `oo*0`

We can rewrite this as

`lim_(x->oo) (ln((2x-3)/(2x+5)))/(1/(2x+1))`

Using L'Hopital's rule

`(d)/(dx) ln((2x-3)/(2x+5)) = (2x+5)/(2x-3) ((2x+5)(2)-(2x-3)(2))/(2x+5)^2=(16)/((2x-3)(2x+5))`

`=lim_(x->oo) (16/((2x-3)(2x+5)))/(-2/(2x+1)^2) = lim_(x->oo) (-8(2x+1)^2)/((2x-3)(2x+5))`

This we can solve by dividing both top and bottom by x^2 to get

=lim_(x->oo) (-8(2+1/x^2)^2)/((2-3/x)(2+5/x))=lim_(x->oo) (-8(2)^2)/(2*2)) = -8

So since the `ln(lim) = -8` that means `e^(ln(lim))=lim=e^-8`

So

`lim_(x->oo) ((2x-3)/(2x+5))^(2x-1) = e^(-8)`

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