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) $$ \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} $$.

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**a) Evaluate $$ \lim_{x \rightarrow 0} \frac{1-\cos (x)}{2 x} $$:**

1. Use the Taylor series expansion for $$\cos(x)$$ around $$x=0$$:
   $$
   \cos(x) = 1 - \frac{x^2}{2} + O(x^4)
   $$
2. Thus, the numerator becomes:
   $$
   1 - \cos(x) \approx \frac{x^2}{2}
   $$
3. Substitute into the limit:
   $$
   \frac{1-\cos(x)}{2x} \approx \frac{\frac{x^2}{2}}{2x} = \frac{x^2}{4x} = \frac{x}{4}
   $$
4. Taking the limit as $$x \to 0$$:
   $$
   \lim_{x \rightarrow 0} \frac{x}{4} = 0
   $$

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**b) Evaluate $$ \lim_{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} $$:**

1. Again, using the Taylor series expansion:
   $$
   1 - \cos(x) \approx \frac{x^2}{2}
   $$
   and note that:
   $$
   \sin(x) \approx x
   $$
2. Substitute the approximations into the limit:
   $$
   \frac{1-\cos(x)}{\sin(x)} \approx \frac{\frac{x^2}{2}}{x} = \frac{x}{2}
   $$
3. Taking the limit as $$x \to 0$$:
   $$
   \lim_{x \rightarrow 0} \frac{x}{2} = 0
   $$

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**c) Evaluate $$ \lim_{x \rightarrow 0} \frac{\sin (7x)}{\sin (5x)} $$:**

1. Use the small-angle approximation:
   $$
   \sin(7x) \approx 7x \quad \text{and} \quad \sin(5x) \approx 5x
   $$
2. Substitute into the limit:
   $$
   \frac{\sin(7x)}{\sin(5x)} \approx \frac{7x}{5x} = \frac{7}{5}
   $$
3. Therefore:
   $$
   \lim_{x \rightarrow 0} \frac{\sin(7x)}{\sin(5x)} = \frac{7}{5}
   $$

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**d) Evaluate $$ \lim_{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} $$:**

1. Factor the denominator:
   $$
   x^2 - 4 = (x-2)(x+2)
   $$
2. Rewrite the limit:
   $$
   \frac{\sin(x-2)}{x^2-4} = \frac{\sin(x-2)}{(x-2)(x+2)}
   $$
3. Split the limit into two parts:
   $$
   \lim_{x \rightarrow 2} \frac{\sin(x-2)}{x-2} \cdot \lim_{x \rightarrow 2} \frac{1}{x+2}
   $$
4. Recognize that:
   $$
   \lim_{x \rightarrow 2} \frac{\sin(x-2)}{x-2} = 1 \quad \text{(by the standard limit } \lim_{u \rightarrow 0} \frac{\sin u}{u}=1 \text{)}
   $$
5. And:
   $$
   \lim_{x \rightarrow 2} \frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4}
   $$
6. Multiply the results:
   $$
   1 \cdot \frac{1}{4} = \frac{1}{4}
   $$

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**Final Answers:**

a) $$0$$  
b) $$0$$  
c) $$\frac{7}{5}$$  
d) $$\frac{1}{4}$$
The limits are: a) 0 b) 0 c) 7/5 d) 1/4

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) \( \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \).

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