- #1
Dell
- 590
- 0
i am given the following, and asked if it converges or diverges
[tex]\int[/tex]dx/ln(x) (from 0-1)
what i did was look for a similar function
i took 1/x which is bigger than 1/lnx,
but 1/x diverges from 0-1
can i split my function up and compare
[tex]\int[/tex]dx/ln(x) (from 0-0.5) + [tex]\int[/tex]dx/ln(x) (from 0.5-1)
now i know that 1/x converges from 0.5-1 so 1/lnx does as well
now i need to find a fuction to compare 1/lnx with from 0-0.5
1/lnx < 1/(x+1)
1/(x+1) converges
therefore 1/lnx does as well
therefore 1/lnx is made up of two converging parts from 0-1 so the whole also converges
is this correct?
[tex]\int[/tex]dx/ln(x) (from 0-1)
what i did was look for a similar function
i took 1/x which is bigger than 1/lnx,
but 1/x diverges from 0-1
can i split my function up and compare
[tex]\int[/tex]dx/ln(x) (from 0-0.5) + [tex]\int[/tex]dx/ln(x) (from 0.5-1)
now i know that 1/x converges from 0.5-1 so 1/lnx does as well
now i need to find a fuction to compare 1/lnx with from 0-0.5
1/lnx < 1/(x+1)
1/(x+1) converges
therefore 1/lnx does as well
therefore 1/lnx is made up of two converging parts from 0-1 so the whole also converges
is this correct?