Question

How do you find the limit of `1+ 9/x` as x approaches `oo`?

Answer 1

`1`

Explanation:

The most important thing to recognize here is that when we have:

`lim_(xrarroo)"constant"/x=0`

Take `lim_(xrarroo)9/x`, for example. As `x` approaches infinity, we get numbers that get increasingly closer to `0`:

`9/9=1" "" "9/90=0.1" "" "9/900=0.01" "" "9/9000=0.001`

So, when we have the function `1+9/x`, just the `9/x` approaches `0`:

`lim_(xrarroo)(1+9/x)=1+0=1`