Question

How do you find the limit of `((h-1)^2+1)/h` as h approaches 0?

Answer 1

`lim_(h->0) ((h-1)^2+1)/h = oo`

Explanation:

Plugging in `0` for `h` gives that:

`lim_(h->0) ((h-1)^2+1)/h = ((0-1)^2+1)/0=2/0`

Since `2/0` equates to the indeterminate form `oo`, we can say that `lim_(h->0) ((h-1)^2+1)/h = oo`.

Answer 2

The limit is divergent.

Explanation:

The whole point of evaluating a limit is that we do not generally look (or care) at the behaviour when `h=0`, just when `h rarr 0`

In this case we have:

`lim_(h rarr 0) ((h-1)^2+1)/h = lim_(h rarr 0) ((h^2-2h+1)+1)/h`
`" " = lim_(h rarr 0) (h^2-2h+2)/h`
`" " = lim_(h rarr 0) (h-2+2/h)`
`" " = lim_(h rarr 0) h - lim_(h rarr 0)2+2lim_(h rarr 0)1/h`

The First limit exists:

`lim_(h rarr 0) h = 0`

The Second limit exists

`lim_(h rarr 0)2 = 2`

However the third limit diverges

`2lim_(h rarr 0)1/h rarr oo`

As `1/h`increases without bound as `h rarr 0`

Hence the initial limit is divergent.