Examine f(x) for continuity at x=1 if f(x)=arctan(2)/(x-1) Select one: a. f(x) is continuous at x=1 b. f(x) has an essential discontinuity at x=1 c. It's not possible to examine the function for continuity d. f(x) has a removable discontinuity at x=1 e. f(x) has a jump discontinuity at x=1

To determine the continuity of the function \( f(x) = \arctan\left(\frac{2}{x-1}\right) \) at \( x = 1 \), we need to analyze the behavior of the function as \( x \) approaches 1.<br /><br />First, let's rewrite the function in a more convenient form:<br />\[ f(x) = \arctan\left(\frac{2}{x-1}\right). \]<br /><br />As \( x \) approaches 1, the denominator \( x-1 \) approaches 0, which makes the argument of the arctangent function approach infinity. This indicates that there is a vertical asymptote at \( x = 1 \).<br /><br />To confirm the type of discontinuity, we can consider the limit of \( f(x) \) as \( x \) approaches 1 from both sides:<br />\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \arctan\left(\frac{2}{x-1}\right) = \frac{\pi}{2}, \]<br />\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \arct2}{x-1}\right) = -\frac{\pi}{2}. \]<br /><br />Since the left-hand limit and the right-hand limit are not equal, there is a jump discontinuity at \( x = 1 \).<br /><br />Therefore, the correct answer is:<br />e. \( f(x) \) has a jump discontinuity at \( x = 1 \).