How do you find the limit of #(1-cosx)/(xsinx)# as #x->0#?

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Answer 1 · 2017-11-12T02:23:05

#lim_(x rarr 0) (1- cosx)/(x sinx) = 1/2#

First of all, since as #x rarr 0#, #sinx rarr 0# also, we can rewrite the denominator as #x^2#.

Hence we need to find: #lim_(x rarr 0) (1- cosx)/(x^2) #

Since this still results in an indeterminate #0/0#, we apply L'Hopital's Rule.

#(d/dx(1-cos x)) / (d/dx(x^2)) = sinx/(2x)#

If we substitute 'approaching zero' as a less formal #1/oo#, we arrive at the expression: #(1/oo)/(2/oo)#.

After cancelling out the infinities, this leaves #1/2#.

Or simply let #sinx =x# again, which gives:

#sinx/(2x) = x/(2x) = 1/2#.

Answer 2 · 2017-11-12T02:42:11

Use #lim_(xrarr0)sinx/x=1#

#((1- cosx))/(x sinx) ((1+cosx))/((1+cosx)) = (1-cos^2x)/(x sinx(1+cosx))#

# = sin^2x/(x sinx(1+cosx))#

# = sinx/x * sinx/sinx * 1/(1+cosx)#

#lim_(x rarr 0) (1- cosx)/(x sinx) = lim_(x rarr 0)(sinx/x * sinx/sinx * 1/(1+cosx))#

# = (1)(1)(1/(1+1))=1/2#