Question

How do you find `lim (3x^2+4)/(x^2-10x+25)` as `x->5`?

Answer 1

See below.

Explanation:

First, plug in `x = 5` to see if it is a problem:

....

`= (3(5)^2+4)/((5)^2-10(5)+25) = "ndef"`, because the denominator is zero :(

We can then see that:

`lim_(x to 5) (3x^2+4)/(x^2-10x+25)`

`= lim_(x to 5) (3x^2+4)/((x-5)^2)`

Clearly this hits `+oo` as `x to 5` !! (NB: It's not a 2-sided limit, which is often an issue.)

As another approach, in these kinda problems, it can be useful to re-state the Origin, so we say that `z = x - 5`, or `x = z+ 5`.

The problem becomes this:

`lim_(z to 0) (3(z+5)^2+4)/((z+5)^2-10(z+5)+25)`

`= lim_(z to 0) (3z^2+30 z + 79)/(z^2)`

`= lim_(z to 0) 3+30 /z + 79/z^2`

Same conclusion :)