# Find the limit. limit goes to zero. f(x)=(cot(x)-(1/x))

`lim_(x->0) (cot(x)-(1/x)) = lim_(x->0) (xcot(x)-1)/x = lim_(x->0) (xcos(x)/sin(x)-1)/x`

`=lim_(x->0)(xcos(x)-sin(x))/(xsin(x))`    Now this is 0/0 and we can use L'Hospital's rule.

`(d)/(dx) (xcos(x)-sin(x)) = cos(x) - xsin(x) - cos(x) = -xsin(x)`

`(d)/(dx) (xsin(x)) = xcos(x) + sin(x)`

`= lim_(x->0) (-xsin(x))/(xcos(x)+sin(x))`    this is also 0/0 so we do it again

`(d)/(dx)...

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`lim_(x->0) (cot(x)-(1/x)) = lim_(x->0) (xcot(x)-1)/x = lim_(x->0) (xcos(x)/sin(x)-1)/x`

`=lim_(x->0)(xcos(x)-sin(x))/(xsin(x))`    Now this is 0/0 and we can use L'Hospital's rule.

`(d)/(dx) (xcos(x)-sin(x)) = cos(x) - xsin(x) - cos(x) = -xsin(x)`

`(d)/(dx) (xsin(x)) = xcos(x) + sin(x)`

`= lim_(x->0) (-xsin(x))/(xcos(x)+sin(x))`    this is also 0/0 so we do it again

`(d)/(dx) (-xsin(x)) = -xcos(x)-sin(x)`

`(d)/(dx) (xcos(x)+sin(x)) = -xsin(x)+cos(x)+cos(x) = 2cos(x)-xsin(x)`

` = lim_(x->0) (-xcos(x)-sin(x))/(2cos(x)-xsin(x))`

now this isn't 0/0 anymore

`= (-0(1)-(0))/(2(1)-0(0)) = 0/2 = 0`