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
For the first question, writing that $\displaystyle \ln(2x)=\ln 2 + \ln x= \left(\frac{\ln 2}{\ln x} + 1\right)\ln x$ may help you.
For the second one, notice that if $\displaystyle x$ tends to 1 from the left, then $\displaystyle 1-x$ tends to 0 from the right, so that the limit you are looking for is in fact the limit of $\displaystyle \ln y$ when $\displaystyle y$ tends to 0 from the right.
Laurent.
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 $\displaystyle \ln (ab)=\ln a+\ln b$... 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 $\displaystyle 1-x$, as $\displaystyle x$ tends to 1 from the right, is $\displaystyle 0^+$ (0, reached by upper values). And the limit of $\displaystyle \ln(y)$ when $\displaystyle y$ tends to 0 from the right is $\displaystyle -\infty$. So, composing these results, the limit of $\displaystyle \ln(1-x)$ when $\displaystyle x$ tends to 1 from the left is $\displaystyle -\infty$.
Is it clearer now?
Laurent.