For `agt0,` let `l=lim_(xto(pi)/2)(a^(cotx)-a^(cosx))/(cotx-cosx)` and `m=lim_(xto-oo)(sqrt(x^(2)+ax))-(sqrt(x^(2)-ax))` then solve it

Answer

To solve the given problem, we need to find the limits \( l \) and \( m \) as defined in the question. Let's break it down step by step.

### Step 1: Finding \( l = \lim_{x \to \frac{\pi}{2}} \frac{a^{\cot x} - a^{\cos x}}{\cot x - \cos x} \)

1. **Substituting the limit**: As \( x \) approaches \( \frac{\pi}{2} \), both \( \cot x \) and \( \cos x \) approach 0. Thus, we have an indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**: Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is of the form \( \frac{0}{0} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

3. **Finding derivatives**:
- Let \( f(x) = a^{\cot x} - a^{\cos x} \)
- Let \( g(x) = \cot x - \cos x \)
- Differentiate \( f(x) \) and \( g(x) \):
- \( f'(x) = a^{\cot x} \cdot \frac{-a^{\cot x} \ln a}{\sin^2 x} - a^{\cos x} \cdot (-\sin x \ln a) \)
- \( g'(x) = -\csc^2 x + \sin x \)

4. **Evaluating the limit**:
- Substitute \( x = \frac{\pi}{2} \) into \( f'(x) \) and \( g'(x) \) and simplify.

5. **Final result for \( l \)**:
- After simplification, we find that \( l = \log a \).

### Step 2: Finding \( m = \lim_{x \to -\infty} \left( \sqrt{x^2 + ax} - \sqrt{x^2 - ax} \right) \)

1. **Rearranging the expression**: Multiply and divide by the conjugate:
\[ m = \lim_{x \to -\infty} \frac{(\sqrt{x^2 + ax} - \sqrt{x^2 - ax})(\sqrt{x^2 + ax} + \sqrt{x^2 - ax})}{\sqrt{x^2 + ax} + \sqrt{x^2 - ax}} \]

2. **Simplifying the numerator**: The numerator becomes:
\[ (x^2 + ax) - (x^2 - ax) = 2ax \]

3. **Simplifying the denominator**: The denominator can be simplified as:
\[ \sqrt{x^2 + ax} + \sqrt{x^2 - ax} \]

4. **Factoring out \( x^2 \)**:
- Factor \( x^2 \) out of the square roots:
\[ \sqrt{x^2(1 + \frac{a}{x})} + \sqrt{x^2(1 - \frac{a}{x})} = |x|\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 - \frac{a}{x}}\right) \]
- Since \( x \to -\infty \), \( |x| = -x \).

5. **Final limit evaluation**:
- Substitute \( x \to -\infty \):
\[ m = \lim_{x \to -\infty} \frac{2ax}{-x(\sqrt{1 + \frac{a}{x}} + \sqrt{1 - \frac{a}{x}})} = \lim_{x \to -\infty} \frac{2a}{-\left(\sqrt{1 + 0} + \sqrt{1 - 0}\right)} = \frac{2a}{-2} = -a \]

### Final Results:
- \( l = \log a \)
- \( m = -a \)