7. Evaluate the following limits: a) \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \) b) \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \) c) \( \lim _{x \rightarrow 0} \frac{\sin (7 x)}{\sin (5 x)} \) d) \( \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \).

Let's evaluate each of the limits step by step: a) To evaluate \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \), we can use the Taylor expansion of \( \cos(x) \) near \( x=0 \): \[ \cos(x) \approx 1 - \frac{x^2}{2} + O(x^4). \] Thus, \[ 1 - \cos(x) \approx \frac{x^2}{2}. \] Substituting this into our limit gives: \[ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{2x} \approx \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{2x} = \lim_{x \rightarrow 0} \frac{x}{4} = 0. \] b) For the limit \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \), again employing the Taylor expansion: \[ 1 - \cos(x) \approx \frac{x^2}{2}, \quad \text{and} \quad \sin(x) \approx x. \] So we rewrite the limit: \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \rightarrow 0} \frac{x}{2} = 0. \] c) Now for \( \lim _{x \rightarrow 0} \frac{\sin (7 x)}{\sin (5 x)} \), we use the small-angle approximation \( \sin(kx) \approx kx \): \[ \lim_{x \rightarrow 0} \frac{\sin(7x)}{\sin(5x)} \approx \lim_{x \rightarrow 0} \frac{7x}{5x} = \frac{7}{5}. \] d) Finally, for \( \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \), notice that both the numerator and the denominator approach \(0\) as \(x \to 2\). Use L'Hôpital's Rule: \[ \frac{d}{dx}[\sin(x - 2)] = \cos(x - 2), \quad \text{and} \quad \frac{d}{dx}[x^2 - 4] = 2x. \] Now apply L'Hôpital's Rule: \[ \lim_{x \rightarrow 2} \frac{\cos(x - 2)}{2x} = \frac{\cos(0)}{2(2)} = \frac{1}{4}. \] In summary, the limits are: a) \( 0 \) b) \( 0 \) c) \( \frac{7}{5} \) d) \( \frac{1}{4} \)