Evaluate the following limits:
`lim_(xrarra)((sinx-sina))/((x-a))`

Answer

To evaluate the limit

\[ \lim_{x \to a} \frac{\sin x - \sin a}{x - a}, \]

we can use the trigonometric identity for the difference of sines. The identity states:

\[ \sin x - \sin a = 2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right). \]

### Step 1: Apply the identity

Using the identity, we rewrite the limit:

\[ \lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \lim_{x \to a} \frac{2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right)}{x - a}. \]

### Step 2: Simplify the expression

Now, we can factor out \(2\) from the limit:

\[ = 2 \lim_{x \to a} \frac{\cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right)}{x - a}. \]

### Step 3: Rewrite the limit

To simplify further, we can rewrite \(x - a\) in terms of \(\sin\left(\frac{x - a}{2}\right)\):

\[ = 2 \lim_{x \to a} \cos\left(\frac{x + a}{2}\right) \cdot \frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}} \cdot \frac{1}{2}. \]

### Step 4: Evaluate the limit

As \(x\) approaches \(a\), \(\frac{x - a}{2}\) approaches \(0\). We know that:

\[ \lim_{u \to 0} \frac{\sin u}{u} = 1. \]

Thus, we have:

\[ \lim_{x \to a} \frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}} = 1. \]

### Step 5: Substitute back into the limit

Now, we can substitute this back into our limit:

\[ = 2 \cdot \cos\left(\frac{a + a}{2}\right) \cdot 1 = 2 \cdot \cos(a). \]

### Final Answer

Therefore, the limit evaluates to:

\[ \lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos(a). \]

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