Evaluate the following limits :
`Lim _( xto a) (tan x - tan a)/(sin a - sin x )`

Answer

To evaluate the limit

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

we start by substituting \(x = a\):

1. **Substitution**:
\[ \tan a - \tan a = 0 \quad \text{and} \quad \sin a - \sin a = 0. \]
This gives us the indeterminate form \(\frac{0}{0}\).

2. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}\) or \(\frac{\pm \infty}{\pm \infty}\), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, \]
provided the limit on the right exists.

3. **Differentiate the Numerator and Denominator**:
- The derivative of the numerator \(\tan x - \tan a\) is:
\[ \frac{d}{dx}(\tan x) = \sec^2 x. \]
- The derivative of the denominator \(\sin a - \sin x\) is:
\[ \frac{d}{dx}(-\sin x) = -\cos x. \]

4. **Rewrite the Limit**:
Now we can rewrite the limit as:
\[ \lim_{x \to a} \frac{\sec^2 x}{-\cos x}. \]

5. **Substitute \(x = a\)**:
Substitute \(x = a\) into the limit:
\[ \frac{\sec^2 a}{-\cos a} = \frac{1/\cos^2 a}{-\cos a} = -\frac{1}{\cos^3 a}. \]

6. **Final Result**:
Therefore, the limit evaluates to:
\[ -\sec^3 a. \]

### Summary of Steps:
1. Substitute \(x = a\) to check for indeterminate form.
2. Apply L'Hôpital's Rule due to \(\frac{0}{0}\) form.
3. Differentiate the numerator and denominator.
4. Rewrite the limit with the derivatives.
5. Substitute \(x = a\) into the new limit.
6. Simplify to find the final result.