Let's analyze and solve the given mathematical expression.

Problem Statement:

The given expression is: $\lim_{x \to 0} \left( \frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \cdot \frac{1}{x} \right).$

Solution:

Step 1: Simplify $a^{x+1}, b^{x+1}, c^{x+1}$

Using the property of exponents, $a^{x+1} = a \cdot a^x$, $b^{x+1} = b \cdot b^x$, and $c^{x+1} = c \cdot c^x$. For small values of $x$, $a^x \approx 1 + x \ln a$ (first-order Taylor expansion).

Thus: $a^{x+1} \approx a(1 + x \ln a) = a + ax \ln a,$ $b^{x+1} \approx b(1 + x \ln b) = b + bx \ln b,$ $c^{x+1} \approx c(1 + x \ln c) = c + cx \ln c.$

Step 2: Substitute these expansions into the numerator

The numerator is: $a^{x+1} - b^{x+1} - c^{x+1}.$

Substituting the expansions: $a^{x+1} - b^{x+1} - c^{x+1} \approx (a + ax \ln a) - (b + bx \ln b) - (c + cx \ln c).$

Simplify: $a^{x+1} - b^{x+1} - c^{x+1} \approx (a - b - c) + x(a \ln a - b \ln b - c \ln c).$

Step 3: Simplify the fraction

The fraction becomes: $\frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \approx \frac{(a - b - c) + x(a \ln a - b \ln b - c \ln c)}{a + b + c}.$

Split into two terms: $\frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c} \approx \frac{a - b - c}{a + b + c} + x \cdot \frac{a \ln a - b \ln b - c \ln c}{a + b + c}.$

Step 4: Multiply by $\frac{1}{x}$

The entire expression is: $\frac{1}{x} \cdot \frac{a^{x+1} - b^{x+1} - c^{x+1}}{a + b + c}.$

Substituting: $\frac{1}{x} \cdot \left( \frac{a - b - c}{a + b + c} + x \cdot \frac{a \ln a - b \ln b - c \ln c}{a + b + c} \right).$

Simplify: $\frac{1}{x} \cdot \frac{a - b - c}{a + b + c} + \frac{a \ln a - b \ln b - c \ln c}{a + b + c}.$

As $x \to 0$, the term $\frac{1}{x} \cdot \frac{a - b - c}{a + b + c}$ diverges unless $a = b + c$. Assuming $a = b + c$, the first term vanishes, leaving: $\lim_{x \to 0} \left( \frac{1}{x} \cdot \frac{a - b - c}{a + b + c} + \frac{a \ln a - b \ln b - c \ln c}{a + b + c} \right) = \frac{a \ln a - b \ln b - c \ln c}{a + b + c}.$

Final Answer:

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

Would you like more details or explanations about specific steps?

5 Related Questions:

1. How is the Taylor expansion of $a^x$ derived for small $x$?
2. What happens if $a \neq b + c$ in this problem?
3. How can L'Hôpital's Rule be applied to limits involving indeterminate forms like this one?
4. What are some common uses of logarithmic expansions in calculus?
5. How can we generalize this result for $n$-term expressions?

Tip:

When solving limits with exponential terms, Taylor expansion and logarithmic approximations are invaluable tools for simplification.