Evaluate `lim_(xto0) (a^(tanx)-a^(sinx))/(tanx-sinx),agt0`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x}, \quad a > 0, \]

we will follow these steps:

### Step 1: Rewrite the limit expression

We start with the limit expression:

\[ \lim_{x \to 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x}. \]

### Step 2: Factor out \(a^{\sin x}\) from the numerator

We can factor \(a^{\sin x}\) out of the numerator:

\[ = \lim_{x \to 0} \frac{a^{\sin x} \left( a^{\tan x - \sin x} - 1 \right)}{\tan x - \sin x}. \]

### Step 3: Evaluate \(a^{\sin x}\) as \(x\) approaches 0

As \(x\) approaches 0, \(\sin x\) approaches 0. Thus,

\[ a^{\sin x} \to a^0 = 1. \]

### Step 4: Simplify the limit

Now we need to evaluate the limit of the remaining part:

\[ \lim_{x \to 0} \frac{a^{\tan x - \sin x} - 1}{\tan x - \sin x}. \]

### Step 5: Use the identity for limits

We can use the identity:

\[ \lim_{u \to 0} \frac{a^u - 1}{u} = \ln a, \]

where \(u = \tan x - \sin x\). To apply this identity, we need to find the limit of \(\tan x - \sin x\) as \(x \to 0\).

### Step 6: Evaluate \(\tan x - \sin x\)

Using Taylor series expansions around \(x = 0\):

\[ \tan x \approx x + \frac{x^3}{3} + O(x^5), \]
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5). \]

Thus,

\[ \tan x - \sin x \approx \left( x + \frac{x^3}{3} \right) - \left( x - \frac{x^3}{6} \right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{x^3}{2}. \]

### Step 7: Substitute back into the limit

Now, substituting back, we have:

\[ \lim_{x \to 0} \frac{a^{\tan x - \sin x} - 1}{\tan x - \sin x} = \lim_{x \to 0} \frac{a^{\frac{x^3}{2}} - 1}{\frac{x^3}{2}}. \]

### Step 8: Apply the identity

As \(x \to 0\), \(\frac{x^3}{2} \to 0\). Thus, we can apply the identity:

\[ \lim_{u \to 0} \frac{a^u - 1}{u} = \ln a, \]

which gives us:

\[ \lim_{x \to 0} \frac{a^{\frac{x^3}{2}} - 1}{\frac{x^3}{2}} = \ln a. \]

### Step 9: Combine results

Putting everything together, we have:

\[ \lim_{x \to 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x} = a^0 \cdot \ln a = 1 \cdot \ln a = \ln a. \]

### Final Answer

Thus, the final answer is:

\[ \ln a. \]