Use l'Hospital's Rule to find the exact value of the limit.
lim 
x→0 
7x − 6x
5x − 4x


To solve the limit using **L'Hospital's Rule**, we first need to verify that the given expression results in an indeterminate form (like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)) when we substitute \(x = 0\). If it does, we can apply L'Hospital's Rule, which involves differentiating the numerator and denominator separately.

The given limit is:

\[
\lim_{x \to 0} \frac{7x - 6x}{5x - 4x}
\]

### Step 1: Check if it forms an indeterminate \(\frac{0}{0}\)

Substitute \(x = 0\) into the numerator and denominator:

- Numerator: \(7^0 - 6^0 = 1 - 1 = 0\)
- Denominator: \(5^0 - 4^0 = 1 - 1 = 0\)

This gives us the indeterminate form \(\frac{0}{0}\), so we can apply L'Hospital's Rule.

### Step 2: Differentiate the numerator and denominator

1. **Numerator**: \(7^x - 6^x\)

Differentiate using the exponential rule: \(\frac{d}{dx}a^x = a^x \ln a\)

\[
\frac{d}{dx}(7^x - 6^x) = 7^x \ln 7 - 6^x \ln 6
\]

2. **Denominator**: \(5^x - 4^x\)

Differentiate using the same rule:

\[
\frac{d}{dx}(5^x - 4^x) = 5^x \ln 5 - 4^x \ln 4
\]

### Step 3: Re-evaluate the limit

Now, we can substitute \(x = 0\) into the derivatives:

- Numerator: \(7^0 \ln 7 - 6^0 \ln 6 = 1 \cdot \ln 7 - 1 \cdot \ln 6 = \ln 7 - \ln 6 = \ln \left(\frac{7}{6}\right)\)
- Denominator: \(5^0 \ln 5 - 4^0 \ln 4 = 1 \cdot \ln 5 - 1 \cdot \ln 4 = \ln 5 - \ln 4 = \ln \left(\frac{5}{4}\right)\)

### Step 4: Final limit

\[
\lim_{x \to 0} \frac{7^x - 6^x}{5^x - 4^x} = \frac{\ln \left(\frac{7}{6}\right)}{\ln \left(\frac{5}{4}\right)}
\]

Thus, the exact value of the limit is:

\[
\boxed{\frac{\ln \left(\frac{7}{6}\right)}{\ln \left(\frac{5}{4}\right)}}
\]

Would you like a more detailed explanation of any steps?

Here are some related questions:
1. How do you apply L'Hospital's Rule to more complex functions?
2. What are the conditions for using L'Hospital's Rule?
3. Can L'Hospital's Rule be used for indeterminate forms other than \(\frac{0}{0}\)?
4. What are some alternatives to L'Hospital's Rule when limits don't result in indeterminate forms?
5. How does exponential differentiation work in other limit problems?

**Tip**: Always verify that the indeterminate form exists before applying L'Hospital's Rule.

Use l'Hospital's Rule to find the exact value of the limit. lim x→0 (7x − 6x) / (5x − 4x)

Learn how to apply L'Hospital's Rule to solve the limit lim x→0 (7x − 6x) / (5x − 4x) with step-by-step differentiation and evaluation of exponential functions. Ideal for students studying Calculus I or those seeking to understand indeterminate forms and limit evaluation.

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