# Why not simplify earlier? 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!

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor 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!

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
Staff Emeritus