Question

lim for x to - infinity of x-log(2^(-x)+1)

Answer 1

\[\lim_{x\rightarrow -\infty}x-\log(2^{-x}+1)\]

Answer 2

it's that how can you handle with it?

Answer 3

oh sorry there's a plus instead of a min in front of the log

Answer 4

i was just writing to see if this was the right thing
i guess it is maybe
\[\lim_{x\rightarrow -\infty}x+\log(2^{-x}+1)\]??

Answer 5

yes it is

Answer 6

sorry browser took a dump

Answer 7

there might be some fancy way to do this that i do not know, but i would think of it like this.
the +1 is clearly irrelevant and if you had
\[x-\log(2^{-x})\] this is the same as
\[x+x\log(2)\] and it is pretty clear that this goes to - infinity as x does

Answer 8

ok that lousy answer bothered me so lets try this
\[\log(2^{-x}+1)=\log(\frac{1}{2^x}+1)\]
\[=\log(\frac{1+2^x}{2^x})=\log(1+2^x)-\log(2^x)=\log(1+2^x)-x\log(2)\]
and so
\[x+\log(2^{-x}+1)=x-x\log(2)+\log(1+2^x)\] now take the limit as x goes to minus infinity

Answer 9

since
\[\log(2)<1\] we know
\[1-\log(2)> 0\] and so
\[\lim_{x\rightarrow -\infty} (1-\log(2))x=-\infty\] and also
\[\lim_{x\rightarrow -\infty} \log(1+2^x)=0\] and therefore we get
\[-\infty\]

Answer 10

thank you very much you have been very helpfull