Can the equation be simplified in the third step if it's indeterminate?

[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!

 

[Mark44]

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.

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

Yet, the solution uses l'Hospital's rule again but it seems simplifiable here at this intermediate step. Any clarification is greatly appreciated!

 

[MathewsMD]

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

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?

 

[vela]

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.