The given expression is:

$\lim_{x \to 0} \left( \frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \cdot \frac{1}{x} \right)$

Step-by-step solution:

1. Focus on the numerator $a^{x+1} - b^{x+1} - c^{x+1}$:

   • When $x \to 0$, all terms $a^{x+1}, b^{x+1}, c^{x+1}$ tend to their respective values at $x=0$, which are $a$, $b$, and $c$. However, we will expand them to capture the effect of $x$ using the exponential approximation: a^{x+1} = a \cdot a^x \approx a \cdot (1 + x \ln a), \quad \text{(since a^x \approx 1 + x \ln a)} Similarly: $b^{x+1} \approx b \cdot (1 + x \ln b), \quad c^{x+1} \approx c \cdot (1 + x \ln c)$
   • Substituting these approximations: $a^{x+1} - b^{x+1} - c^{x+1} \approx a (1 + x \ln a) - b (1 + x \ln b) - c (1 + x \ln c)$ Simplify: $a^{x+1} - b^{x+1} - c^{x+1} \approx (a - b - c) + x (a \ln a - b \ln b - c \ln c)$

2. Substitute this result into the limit: Substitute the numerator and divide by $(a + b + c)$: $\frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \approx \frac{(a - b - c) + x (a \ln a - b \ln b - c \ln c)}{a + b + c}$

3. Multiply by $\frac{1}{x}$: After multiplying by $\frac{1}{x}$, the term $\frac{a - b - c}{a + b + c}$ vanishes as $x \to 0$, and the remaining term becomes: $\frac{1}{x} \cdot \frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \approx \frac{a \ln a - b \ln b - c \ln c}{a + b + c}$

4. Final Limit: $\lim_{x \to 0} \left( \frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \cdot \frac{1}{x} \right) = \frac{a \ln a - b \ln b - c \ln c}{a + b + c}$

Final Answer:

$\boxed{\frac{a \ln a - b \ln b - c \ln c}{a + b + c}}$

Would you like further clarification or a detailed explanation of any specific part? 😊

Related Questions:

1. What is the approximation $a^x$ for small $x$?
2. How does L'Hôpital's Rule apply to exponential functions in limits?
3. Can this problem be solved without logarithmic expansion?
4. How would the answer change if $c = 0$?
5. What happens if $a = b = c$ in the given limit?

Tip: When solving exponential limits, approximations like $e^x \approx 1 + x$ (for small $x$) often simplify calculations.