1 - cos h
lim h -0 --------- = 0 how ?
h

Question

1 - cos h
 lim h ->0      ---------  =  0       how ?
                             h

anonymous · Answer

is there anyone to help me.. :) :)

aravindg · Answer

attachment

lgbasallote · Answer

do you know how to use L'Hospital's Rule? And this is a Math question NOT physics

aravindg · Answer

1-cosx is 2sin^2x/2

experimentx · Answer

$ 1-\cos x = 1 - \cos^2(x/2)  + \sin^2(x/2) = 2 \sin^ (x/2)$  <--- change cosine in to half angle formula

anonymous · Answer

@AravindG  g i know it is 0 but how :) 
@lgbasallote  no i really don't know ..

lgbasallote · Answer

wow so many possible answers o.O all i know is l'hospital lol

anonymous · Answer

@experimentX  i got u till now further plz :) 
@lgbasallote  it's alright :)

experimentx · Answer

sinx/ x = 1 <--- make it to this form. since there is square in the top

x/2 * sin^2x/2/(x/2)^2 <----now you can solve.

anonymous · Answer

from where "x/2 * sin^2x/2/(x/2)^2" came ?

anonymous · Answer

@experimentX    from where      " x/2 * sin^2x/2/(x/2)^2" came ?

experimentx · Answer

try balancing things yourself.

anonymous · Answer

@experimentX  i got i it thanks alot :) 
@AravindG  thanks u too :)

aravindg · Answer

welcomess