Question

How do you evaluate the limit `x-2` as x approaches `1`?

Answer 1

Plug in the value `1` into `x`.

Thus:

`lim_(x->1) x-2=1-2=-1`

It only starts to get complex when you plug in the value that `x` approaches, and you're left with an indeterminate form.

(ie. `0/0` or `oo/oo` or `oo^oo`) . . . the list goes on.