lim _(x arrow 2) operatorname(tg) (pi)/(4)(2-x)

To find the limit \( \lim_{x \rightarrow 2} \operatorname{tg} \frac{\pi}{4}(2-x) \), we can simplify the expression inside the tangent function.<br /><br />First, let's rewrite the expression inside the tangent function:<br />\[ \frac{\pi}{4}(2-x) \]<br /><br />As \( x \) approaches 2, \( 2-x \) approaches 0. Therefore, we have:<br />\[ \lim_{x \rightarrow 2} \operatorname{tg} \left( \frac{\pi}{4}(2-x) \right) \]<br /><br />Now, let's evaluate the limit:<br />\[ \lim_{x \rightarrow 2} \operatorname{tg} \left( \frac{\pi}{4}(2-x) \right) = \operatorname{tg} \left( \lim_{x \rightarrow 2} \frac{\pi}{4}(2-x) \right) \]<br /><br />Since \( \frac{\pi}{4}(2-x) \) approaches 0 as \( x \) approaches 2, we have:<br />\[ \lim_{x \rightarrow 2} \operatorname{tg} \left( \frac{\pi}{4}(2-x) \right) = \operatorname{tg}(0) \]<br /><br />The tangent of 0 is 0, so the limit is:<br />\[ \lim_{x \rightarrow 2} \operatorname{tg} \left( \frac{\pi}{4}(2-x) \right) = 0 \]<br /><br />Therefore, the correct answer is 0.