Asked by Sarah
Limit as x approaches zero (1/x-1/tanx)
Answers
Answered by
Steve
massage the form to
1/x - cosx/sinx
= (cosx-x)/(x*sinx)
and apply l'Hospital's Rule
1/x - cosx/sinx
= (cosx-x)/(x*sinx)
and apply l'Hospital's Rule
Answered by
Sarah
That is where I am confused ...
Answered by
Steve
Oops. A typo. Should be
(x*cosx-sinx)/(x*sinx)
just take derivatives, top and bottom. The limit is the same as for
(-x*sinx+cosx-cosx)/(sinx + xcosx)
= (-x sinx)/(sinx + x cosx)
still -> 0/0, so do it again:
(-sinx - x*cosx)/(cosx + cosx - x*sinx)
-> 0/2 = 0
(x*cosx-sinx)/(x*sinx)
just take derivatives, top and bottom. The limit is the same as for
(-x*sinx+cosx-cosx)/(sinx + xcosx)
= (-x sinx)/(sinx + x cosx)
still -> 0/0, so do it again:
(-sinx - x*cosx)/(cosx + cosx - x*sinx)
-> 0/2 = 0