How do I find the limit as x goes to +infinity for (ln 2x)/(ln 3x)?
Also, how do I find the limit as x approaches 1 from the negative direction for ln (1-x)?
Thank You.
Jeannine
Thanks. I get the first one.
For the second, I'm still unclear. I notice that:
ln(1-x) = ln 1 - ln x = 0 - ln x = -ln x.
Does this help? How do I find the limit as x approaches 1 from the negative direction for -ln x?
Sorry for my clouded mind!
Jeannine
No you don't. You have ... Logarithms turns multiplications into additions (by the way, log was first introduced to simplify computations, since additions are easier).
What I suggested in my first post may be called "composition of limits". The limit of , as tends to 1 from the right, is (0, reached by upper values). And the limit of when tends to 0 from the right is . So, composing these results, the limit of when tends to 1 from the left is .
Is it clearer now?
Laurent.