Key Concepts and Formulas

• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Standard Limit: $\mathop {\lim }\limits_{x \to 0} \frac{e^{kx} - 1}{kx} = 1$. This can be derived using L'Hôpital's Rule or Taylor series expansion.
• Limit Properties: The limit of a product is the product of the limits, and the limit of a sum/difference is the sum/difference of the limits, provided these limits exist.

Step-by-Step Solution

The problem asks us to find the value of $a - 2b$, given that the limit $L = \mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$ exists and is equal to $b$.

Step 1: Analyze the indeterminate form of the limit. As $x \to 0$, the numerator approaches $a(0) - (e^{4(0)} - 1) = 0 - (1 - 1) = 0$. The denominator approaches $a(0)(e^{4(0)} - 1) = 0(1 - 1) = 0$. Thus, the limit is of the indeterminate form $\frac{0}{0}$. This suggests we can use L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule for the first time. Let $f(x) = ax - (e^{4x} - 1)$ and $g(x) = ax(e^{4x} - 1)$. Then $f'(x) = a - 4e^{4x}$. And $g'(x) = a(e^{4x} - 1) + ax(4e^{4x}) = ae^{4x} - a + 4axe^{4x}$. Applying L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 0} \frac{a - 4e^{4x}}{ae^{4x} - a + 4axe^{4x}}$

Step 3: Analyze the indeterminate form after the first application of L'Hôpital's Rule. As $x \to 0$, the numerator approaches $a - 4e^{4(0)} = a - 4$. The denominator approaches $ae^{4(0)} - a + 4a(0)e^{4(0)} = a - a + 0 = 0$. For the limit $L$ to exist (and be equal to $b$), the numerator must also approach 0 when the denominator approaches 0. Therefore, we must have $a - 4 = 0$.

Step 4: Determine the value of 'a'. From the condition $a - 4 = 0$, we get $a = 4$.

Step 5: Re-evaluate the limit with $a=4$ using the standard limit form. Substitute $a=4$ into the original limit expression: $L = \mathop {\lim }\limits_{x \to 0} \frac{4x - (e^{4x} - 1)}{4x(e^{4x} - 1)}$ We can rewrite this as: $L = \mathop {\lim }\limits_{x \to 0} \frac{1}{4x} \left( \frac{4x - (e^{4x} - 1)}{e^{4x} - 1} \right)$ $L = \mathop {\lim }\limits_{x \to 0} \frac{1}{4x} \left( \frac{4x}{e^{4x} - 1} - \frac{e^{4x} - 1}{e^{4x} - 1} \right)$ $L = \mathop {\lim }\limits_{x \to 0} \frac{1}{4x} \left( \frac{4x}{e^{4x} - 1} - 1 \right)$ $L = \mathop {\lim }\limits_{x \to 0} \left( \frac{1}{e^{4x} - 1} - \frac{1}{4x} \right)$ This still looks problematic. Let's go back to the expression after the first L'Hopital's Rule with $a=4$.

Step 5 (Revised): Re-evaluate the limit with $a=4$ after the first application of L'Hôpital's Rule. With $a=4$, the limit after the first application of L'Hôpital's Rule becomes: $L = \mathop {\lim }\limits_{x \to 0} \frac{4 - 4e^{4x}}{4e^{4x} - 4 + 16xe^{4x}}$ This is still of the form $\frac{0}{0}$ since the numerator is $4 - 4(1) = 0$ and the denominator is $4(1) - 4 + 16(0)(1) = 0$. We need to apply L'Hôpital's Rule again.

Step 6: Apply L'Hôpital's Rule for the second time. Let $f_1(x) = 4 - 4e^{4x}$ and $g_1(x) = 4e^{4x} - 4 + 16xe^{4x}$. Then $f_1'(x) = -16e^{4x}$. And $g_1'(x) = 4(4e^{4x}) - 0 + 16(1 \cdot e^{4x} + x \cdot 4e^{4x}) = 16e^{4x} + 16e^{4x} + 64xe^{4x} = 32e^{4x} + 64xe^{4x}$. Applying L'Hôpital's Rule again: $L = \mathop {\lim }\limits_{x \to 0} \frac{-16e^{4x}}{32e^{4x} + 64xe^{4x}}$

Step 7: Evaluate the limit after the second application. As $x \to 0$, the numerator approaches $-16e^{4(0)} = -16$. The denominator approaches $32e^{4(0)} + 64(0)e^{4(0)} = 32(1) + 0 = 32$. So, the limit is: $L = \frac{-16}{32} = -\frac{1}{2}$ We are given that the limit exists and is equal to $b$. Therefore, $b = -\frac{1}{2}$.

Step 8: Calculate the value of $a - 2b$. We found $a = 4$ and $b = -\frac{1}{2}$. $a - 2b = 4 - 2\left(-\frac{1}{2}\right)$ $a - 2b = 4 - (-1)$ $a - 2b = 4 + 1$ $a - 2b = 5$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
• Algebraic Errors during differentiation: Carefully differentiate the numerator and denominator. For the denominator $ax(e^{4x}-1)$, remember to use the product rule.
• Forgetting the condition for existence: The fact that the limit exists is crucial. It implies that if the denominator goes to zero, the numerator must also go to zero, leading to the condition $a-4=0$.
• Using the standard limit form correctly: The limit $\mathop {\lim }\limits_{x \to 0} \frac{e^{kx} - 1}{kx} = 1$ can be a shortcut. Rewriting the original expression: $\mathop {\lim }\limits_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)} = \mathop {\lim }\limits_{x \to 0} \frac{1}{ax} \left( \frac{ax - (e^{4x} - 1)}{e^{4x} - 1} \right)$ $= \mathop {\lim }\limits_{x \to 0} \frac{1}{ax} \left( \frac{ax}{e^{4x} - 1} - 1 \right)$ $= \mathop {\lim }\limits_{x \to 0} \left( \frac{1}{e^{4x} - 1} - \frac{1}{ax} \right)$ This form is not directly resolvable. A better approach with standard limits is: $\mathop {\lim }\limits_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)} = \mathop {\lim }\limits_{x \to 0} \frac{ax - (e^{4x} - 1)}{a} \cdot \frac{1}{x(e^{4x} - 1)}$ $= \mathop {\lim }\limits_{x \to 0} \frac{ax - (e^{4x} - 1)}{a} \cdot \frac{1}{x^2 \frac{e^{4x} - 1}{x}}$ Using Taylor series: $e^{4x} - 1 = 4x + \frac{(4x)^2}{2!} + \dots = 4x + 8x^2 + \dots$ Numerator: $ax - (4x + 8x^2 + \dots) = (a-4)x - 8x^2 - \dots$ Denominator: $ax(4x + 8x^2 + \dots) = 4ax^2 + 8ax^3 + \dots$ The limit is $\mathop {\lim }\limits_{x \to 0} \frac{(a-4)x - 8x^2 - \dots}{4ax^2 + 8ax^3 + \dots}$. For the limit to exist, the term with the lowest power of $x$ in the numerator must match the term with the lowest power of $x$ in the denominator, or be of higher order. If $a-4 \neq 0$, the numerator's lowest power is $x^1$, while the denominator's lowest power is $x^2$. The limit would be $\infty$. So, $a-4 = 0 \implies a=4$. Now the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{-8x^2 - \dots}{4(4)x^2 + \dots} = \mathop {\lim }\limits_{x \to 0} \frac{-8x^2}{16x^2} = \frac{-8}{16} = -\frac{1}{2}$. This confirms $b = -1/2$.

Summary

The problem involves evaluating a limit that initially appears to be in an indeterminate form. We first established that for the limit to exist, the numerator must also tend to zero as the denominator tends to zero, which allowed us to determine the value of $a$. Subsequently, after substituting the value of $a$, we applied L'Hôpital's Rule a second time to evaluate the limit and find the value of $b$. Finally, we computed the required expression $a - 2b$.

The final answer is $\boxed{5}$.