Question

Question #06261

Answer 1

It is because `lim_(x->oo)(1/x)=0`

Explanation:

I assume the question is written as:
`lim_(x->oo)(6/x)-3`

We can first factor out the `6` like so:
`lim_(x->oo)(6/x)-3`
`=lim_(x->oo)(6*1/x)-3`
`=lim_(x->oo)(6)*lim_(x->oo)(1/x)-3`
`=6*lim_(x->oo)(1/x)-3`

Then, we just solve for:

`lim_(x->oo)(1/x)`

Recall that the limit of `1/x` as `x` approaches `oo` is `0`.
This is because, for bigger and bigger values of `x`, `1/x` becomes smaller and smaller.

graph{1/x [-4.25, 15.75, -0.56, 9.44]}

You can see on the above graph, that as you go further along the x-axis, the closer the graph gets to `0`.

So, the limit of this graph as `x` approaches `oo` is `0`.

Now, we know that:
`lim_(x->oo)(1/x)=0`

So:

`6*lim_(x->oo)(1/x)-3`

`=6*0-3`

`=-3`