To solve the given limit problem step by step, we will evaluate:

$\lim_{x \to 0} \frac{6^x - 3^x - 2^x + 1}{x^2}.$

Step 1: Substitute $x = 0$ directly into the expression.

If we substitute $x = 0$: $\text{Numerator: } 6^0 - 3^0 - 2^0 + 1 = 1 - 1 - 1 + 1 = 0.$ $\text{Denominator: } x^2 = 0^2 = 0.$

This gives the indeterminate form $\frac{0}{0}$. Hence, we need to use L'Hôpital's Rule, which states: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \quad \text{if the limit exists}.$

Step 2: Apply L'Hôpital's Rule (differentiate numerator and denominator).

Differentiate the numerator:

The numerator is: $f(x) = 6^x - 3^x - 2^x + 1.$ Differentiating term by term: $f'(x) = 6^x \ln(6) - 3^x \ln(3) - 2^x \ln(2).$

Differentiate the denominator:

The denominator is: $g(x) = x^2.$ Differentiating: $g'(x) = 2x.$

After applying L'Hôpital's Rule: $\lim_{x \to 0} \frac{6^x - 3^x - 2^x + 1}{x^2} = \lim_{x \to 0} \frac{6^x \ln(6) - 3^x \ln(3) - 2^x \ln(2)}{2x}.$

Step 3: Apply L'Hôpital's Rule again (since this still results in $\frac{0}{0}$).

Differentiate the numerator again:

$f'(x) = 6^x \ln(6) - 3^x \ln(3) - 2^x \ln(2).$ Differentiating again: $f''(x) = 6^x (\ln(6))^2 - 3^x (\ln(3))^2 - 2^x (\ln(2))^2.$

Differentiate the denominator again:

$g'(x) = 2x \quad \implies \quad g''(x) = 2.$

After applying L'Hôpital's Rule again: $\lim_{x \to 0} \frac{6^x \ln(6) - 3^x \ln(3) - 2^x \ln(2)}{2x} = \lim_{x \to 0} \frac{6^x (\ln(6))^2 - 3^x (\ln(3))^2 - 2^x (\ln(2))^2}{2}.$

Step 4: Evaluate the limit as $x \to 0$.

As $x \to 0$:

• $6^x \to 1$,
• $3^x \to 1$,
• $2^x \to 1$.

Substitute these values into the differentiated numerator: $f''(0) = 1 \cdot (\ln(6))^2 - 1 \cdot (\ln(3))^2 - 1 \cdot (\ln(2))^2.$

Substitute into the denominator $g''(x) = 2$: $\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{(\ln(6))^2 - (\ln(3))^2 - (\ln(2))^2}{2}.$

Final Answer:

$\boxed{\frac{(\ln(6))^2 - (\ln(3))^2 - (\ln(2))^2}{2}}$

Would you like further clarification or simplifications? 😊
Here are some follow-up questions related to this solution:

1. What is L'Hôpital's Rule and when can it be applied?
2. How do logarithmic properties simplify exponential differentiation?
3. Can we solve this limit using series expansions instead of L'Hôpital's Rule?
4. Why do higher-order derivatives help resolve indeterminate forms like $\frac{0}{0}$?
5. What are some other techniques for solving limits apart from L'Hôpital's Rule?

Tip: Always check for simple substitutions before applying advanced methods like L'Hôpital's Rule.