The value of `lim_(x->0) (x-sin x)/(x-tan x)` has to be determined.

If x is substituted with 0, the value of the expression is 0/0 which is indeterminate. This allows the use of l'Hopital's rule with the substitution of the numerator and denominator with their derivatives.

=> `lim_(x->0) (1-cos x)/(1-sec^2 x)`

=> `lim_(x->0) (1-cos x)/((1-sec x)(1+sec x))`

=> `lim_(x->0) ((1-cos x)(cos^2x))/((cos x-1)(cos x+1))`

=> `lim_(x->0) -(cos^2x)/(cos x+1)`

substitute x = 0

=> -1/2

**The value of **`lim_(x->0) (x-sin x)/(x-tan x) = -1/2`

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