Saturday 23 January 2016

calculus - Computing $limlimits_{xtoinfty}ln(x)cdot ln(1-e^{-x})$

Limit as $x$ approaches infinity of $\ln(x)\cdot \ln(1-e^{-x})$:
$$
\lim_{x\to\infty}\ln(x)\cdot \ln(1-e^{-x})
$$



The only thing I can think to do is rewrite the ln(x) on the bottom as $(lnx)^{-1}$ and use L’Hôpital’s rule, but I’ve done two iterations now and it keeps getting back to the 0/0 or infinity*0 indeterminate case. Any help on how to proceed will be much appreciated! Thanks

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...