# Why not simplify earlier?

1. Oct 31, 2013

### MathewsMD

Why can't the equation be simplified in the third step where it specifies that the equation is indeterminate? You could essentially have ln1/(ln1+1-1) if you substitute values in here. This then simplifies to just ln1/ln1 and then they would cancel to equal 1. Yet, the solution uses l'Hospital's rule again but it seems simplifiable here at this intermediate step. Any clarification is greatly appreciated!

2. Oct 31, 2013

### Staff: Mentor

No, because ln1 = 0. The fraction would simplify to 0/(0 + 1 - 1), so you're at the indeterminate form [0/0].

3. Oct 31, 2013

### MathewsMD

But is not the same as 5/(5+1-1) = 5/5 = 1 or x/(x+1-1) = x/x = 1, where the numerator and denominator become the same? Are we not allowed to do this since these are limits and not exactly these values or is there another reason?

4. Oct 31, 2013

### vela

Staff Emeritus
No, it's not the same. You can say $x/x=1$ only when $x\ne 0$, otherwise, it's undefined.

If a limit results in 0/0, it's indeterminate, and you need to do some work to figure out if the limit exists.