2. (4 points) What types of discontinuities does the following function have? Support your answer with a limit for each \( x \)-value you found. \[ f(x)=\frac{2 x-1}{\left(4 x^{2}-1\right)}=\frac{2 x-1}{(2 x-1)(2 x+1)} \]

The function \( f(x) = \frac{2 x-1}{4 x^{2}-1} \) has discontinuities when the denominator equals zero. This occurs at \( x = \pm \frac{1}{2} \). Specifically, at \( x = \frac{1}{2} \) both the numerator and denominator approach zero, indicating a removable discontinuity. The limit is: \[ \lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}} \frac{2x - 1}{(2x-1)(2x+1)} = \lim_{x \to \frac{1}{2}} \frac{1}{2x + 1} = \frac{1}{2 \cdot \frac{1}{2} + 1} = \frac{1}{2}. \] For \( x = -\frac{1}{2} \), only the denominator goes to zero, signifying an infinite discontinuity. The limit here is: \[ \lim_{x \to -\frac{1}{2}} f(x) = \lim_{x \to -\frac{1}{2}} \frac{2x - 1}{(2x-1)(2x+1)} \text{ which approaches } \infty. \] Thus, we observe a removable discontinuity at \( x = \frac{1}{2} \) and an infinite discontinuity at \( x = -\frac{1}{2} \).