lim as x - 3+ ln(x-3)

Question

lim as x -> 3+     ln(x-3)

anonymous · Answer

that be $-\infty$

konradzuse · Answer

That's what Wolfman said, how exactly do we figure that out?  My textbook was kinda vague....

anonymous · Answer

as $x\to 3^+$ we have $x-3\to 0^+$ and and the log decreases to minus infinity as the input goes to zero

anonymous · Answer

one explanation would come from the graph of $y=\log(x)$ another would be to say $y=\log(x)\iff e^y=x$ and of course $e^y>0$ for all $y$  and $e^y$ goes to zero only if $y$ goes to minus infinity

konradzuse · Answer

http://www.wolframalpha.com/input/?i=lim+as+x+approaches+3%2B++of+ln%28x-3%29 I see here it drops down infinitely, so I guess that's why it's -infinity.

konradzuse · Answer

brb

anonymous · Answer

don't be confused by wolfram's graph. that is a real and complex valued function, but you are only working with real numbers. here is the real graph http://www.wolframalpha.com/input/?i=y+%3D+log%28x%29

konradzuse · Answer

ok I see how it's going from the left to 0 it goes to neg infinity...  What does the x-3 mean in terms of the graph?  -3?

anonymous · Answer

the difference between the graph of $y=f(x)$ and $y=f(x-3)$ is that the second on is shifted to the right 3 units

anonymous · Answer

there is a tab in wolfram plot where you can select "real valued plot" to get rid of the annoying complex part

konradzuse · Answer

Alrighty, that's what I figured.... thanks... Still a little confused on this one I guess... The next question http://www.wolframalpha.com/input/?i=lim+as+x+approaches+7-++of+ln%28x%5E2%288-x%29%29 Makes sense if you plug in 7 for x, but this one just doesn't... if we plug in 3 for this one it should be ln(0) which = 1 right? I'm still confused :(.

konradzuse · Answer

gtg now bbl :P thanks Satellite.