Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\tan x}}} \right) \cr
& {\text{If we evaluate the limit we obtain }}\infty - \infty ,{\text{ so we cannot apply}} \cr
& {\text{the power rule}}{\text{. It is necessary rewrite the function }}\frac{1}{x} - \frac{1}{{\tan x}} \cr
& {\text{Recall that }}\frac{a}{b} - \frac{c}{d}{\text{ = }}\frac{{ad - bc}}{{bd}}{\text{. Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\tan x}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\tan x - x}}{{x\tan x}}} \right) \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\tan x - x}}{{x\tan x}}} \right) = \frac{{\tan 0 - 0}}{{0\tan 0}} = \frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\tan x - x}}{{x\tan x}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {\tan x - x} \right]}}{{\frac{d}{{dx}}\left[ {x\tan x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sec }^2}x - 1}}{{x\left( {{{\sec }^2}x} \right) + \tan x}} \cr
& {\text{Evaluate the limit when }}x \to {0^ + } \cr
& = \frac{{{{\sec }^2}\left( 0 \right) - 1}}{{0\left( {{{\sec }^2}\left( 0 \right)} \right) + \tan 0}} = \frac{0}{0} \cr
& {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sec }^2}x - 1}}{{x\left( {{{\sec }^2}x} \right) + \tan x}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left[ {{{\sec }^2}x - 1} \right]}}{{\frac{d}{{dx}}\left[ {x\left( {{{\sec }^2}x} \right) + \tan x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2\sec x\left( {\sec x\tan x} \right) - 0}}{{x\left( {2\sec x} \right)\left( {\sec x\tan x} \right) + {{2\sec }^2}x}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2{{\sec }^2}x\tan x}}{{2{{\sec }^2}x\tan x + {{2\sec }^2}x}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2\tan x}}{{2\tan x + 2}} \cr
& {\text{Evaluate the limit when }}x \to {0^ + } \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2\tan x}}{{2\tan x + 2}} = \frac{{2\tan \left( 0 \right)}}{{2\tan \left( 0 \right) + 2}} = \frac{0}{2} = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\tan x}}} \right) = 0 \cr} $$
