Question

How do you find the limit of `ln((2x)/(x+1))` as x approaches infinity?

Answer 1

`= ln 2`

Explanation:

`lim_(x to oo) ln((2x)/(x+1))`

as `ln` is continuous we can say:
`=ln ( lim_(x to oo) (2x)/(x+1) )`

`=ln ( lim_(x to oo) (2)/(1+1/x) )`

and as `lim_(x to oo) 1/x = 0`

`=ln ((2)/(1+0) ) = ln 2`