Evaluate `lim_(xto0)((ab)^(x)-a^(x)-b^(x)+1)/(x^(2))`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{(ab)^x - a^x - b^x + 1}{x^2}, \]

we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression in the limit:

\[ (ab)^x - a^x - b^x + 1 = e^{x \ln(ab)} - e^{x \ln a} - e^{x \ln b} + 1. \]

### Step 2: Use Taylor expansion
Using the Taylor expansion of \(e^x\) around \(x=0\):

\[ e^u \approx 1 + u + \frac{u^2}{2} + O(u^3), \]

we can expand each term:

1. \( (ab)^x = e^{x \ln(ab)} \approx 1 + x \ln(ab) + \frac{(x \ln(ab))^2}{2} + O(x^3) \)
2. \( a^x = e^{x \ln a} \approx 1 + x \ln a + \frac{(x \ln a)^2}{2} + O(x^3) \)
3. \( b^x = e^{x \ln b} \approx 1 + x \ln b + \frac{(x \ln b)^2}{2} + O(x^3) \)

### Step 3: Substitute the expansions
Substituting these expansions into the limit expression, we get:

\[ \begin{align*} (ab)^x - a^x - b^x + 1 &\approx \left(1 + x \ln(ab) + \frac{(x \ln(ab))^2}{2}\right) - \left(1 + x \ln a + \frac{(x \ln a)^2}{2}\right) \\ &\quad - \left(1 + x \ln b + \frac{(x \ln b)^2}{2}\right) + 1. \end{align*} \]

### Step 4: Simplify the expression
Simplifying the above expression, we have:

\[ x \ln(ab) - x \ln a - x \ln b + \frac{(x \ln(ab))^2}{2} - \frac{(x \ln a)^2}{2} - \frac{(x \ln b)^2}{2}. \]

Notice that \( \ln(ab) = \ln a + \ln b \), so the linear terms cancel out:

\[ \frac{(x \ln(ab))^2}{2} - \frac{(x \ln a)^2}{2} - \frac{(x \ln b)^2}{2} = \frac{x^2}{2} \left((\ln(ab))^2 - (\ln a)^2 - (\ln b)^2\right). \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:

\[ \lim_{x \to 0} \frac{\frac{x^2}{2} \left((\ln(ab))^2 - (\ln a)^2 - (\ln b)^2\right)}{x^2} = \frac{1}{2} \left((\ln(ab))^2 - (\ln a)^2 - (\ln b)^2\right). \]

### Step 6: Evaluate the limit
We know that:

\[ \ln(ab) = \ln a + \ln b, \]

so

\[ (\ln(ab))^2 = (\ln a + \ln b)^2 = (\ln a)^2 + 2 \ln a \ln b + (\ln b)^2. \]

Thus,

\[ (\ln(ab))^2 - (\ln a)^2 - (\ln b)^2 = 2 \ln a \ln b. \]

### Final Step: Conclusion
Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{(ab)^x - a^x - b^x + 1}{x^2} = \frac{1}{2} \cdot 2 \ln a \ln b = \ln a \ln b. \]

So the final answer is:

\[ \boxed{\ln a \ln b}. \]