Limits Homework question?

Question

Ive got this limits problem im stuck on. The actual limits part is easy, the part im stuck on is the algebra.

lim (lne^(2f+1) - e^(lm6f) + e^(f-1) - 4lnf)

f->1

variables: l, m, f

this is what ive done:

plugged in f as 1: (lne^(2(1)+1) - e^(lm6(1)) - 4ln1)

started solving(thinking they are legal moves)

2+1 - e^(lm6) -4ln1

3 - e^(6lm) - 4(e^0)

-1 - e^(6lm)

ln(-1) + ln(-e^6lm)

ln(-1) - 6lm

and this is where im stuck, not sure if this is as far as it goes, if i messed up, or i am missing something. Please help, ill be checking this in the morning but after around 10 am the answers wont really help anymore.

Answer 1

first of all you made a mistake as f->1  e^(f-1) goes to 1 not zero

{e to the zero power=1} Then the next mistake I see is 4ln1 is zero, not 4 {ln1 =0 so 4ln1=0} So we are at 4-e^(6lm) which is the answer.

I'm not sure why you think you can take the ln of this and then reduce it....you can only do that in an equation where you do it to both sides of the equal sign. Hence, the final answer is 4 - e^(6lm)

ANSWER