Find the limit of (5^x - 6^x)/(7^x - 8^x) as x tends to infinity.

Hello!

Consider `x->+oo` and `x->-oo` separately.

For `x->+oo,` we have an indeterminacy of the type `oo/oo.` To resolve it, divide all terms by the most rapid increasing one, `8^x:`

`((5/8)^x-(6/8)^x)/((7/8)^x-1),`  `x->+oo.`

`(5/8)^x,` `(6/8)^x` and `(7/8)^x` tend to zero as `x` tends to `+oo,` and `(5/8)^xlt(6/8)^x.`

So we obtain `(-0)/(-1) =` +0. This is the limit as `x->+oo.`

For `x->-oo,` there is  `0/0`  and the "main" term is `5^x,` divide by it:

`(1-(6/5)^x)/((7/5)^x-(8/5)^x),`  `x->-oo.`

`(6/5)^x,` `(7/5)^x,` `(8/5)^x` tend to zero as x tends to `-oo` and `(7/5)^xgt(8/5)^x.`

So the result is `(1)/(+0)=+oo.` This is the limit for `x->-oo.`

The answer(s):

`lim_(x->+oo) (5^x-6^x)/(7^x-8^x)=+0,`

`lim_(x->-oo) (5^x-6^x)/(7^x-8^x)=+oo.`