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Topic:
little calculus problem
Hey everybody, can any one find the value of this limit : lim (ln(x) + ln(csc(x))) x ---> 0 ?
Jun 7, 2007
9:40 PM
... = lim ln(x/sin(x)) = ln 1 = 0.
Jun 9, 2007
1:47 AM
hmmmm
...= lim x ln(x/sin(x))......
Jun 25, 2007
5:56 PM
lim[lnx+ln(cscx)]=lim[ln(x*cscx)]=lim[ln(x/sinx)] (1)
If u=x/sinx, then limu=lim(x/sinx)=lim(1/cosx)=1 (De L'Hospital)
As such, (1) gives us: lim[ln(x/sin)]=lim(lnu)=ln1=o, when u->1. So lim[lnx+ln(cscx)]=0
Oct 26, 2008
7:14 PM