Key Concepts and Formulas

• Limit of a Function: The limit of a function $f(x)$ as $x$ approaches $a$, denoted by $\mathop {\lim }\limits_{x \to a} f(x)$, is the value that $f(x)$ approaches as $x$ gets arbitrarily close to $a$.
• Standard Limit: The fundamental limit $\mathop {\lim }\limits_{y \to 0} \frac{e^y - 1}{y} = 1$. This will be crucial for evaluating the given limit.
• Indeterminate Forms: An indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$, arises when direct substitution into a limit expression results in an undefined expression. These forms require further manipulation (like L'Hôpital's Rule or Taylor series expansion) to evaluate the limit.
• Taylor Series Expansion: The Taylor series expansion of $e^y$ around $y=0$ is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$. This is useful for approximating functions near a point and evaluating limits.

Step-by-Step Solution

We are given the limit: $\beta=\mathop {\lim }\limits_{x \to 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}$

Step 1: Analyze the indeterminate form. As $x \to 0$, the numerator becomes $\alpha(0) - (e^{3(0)} - 1) = 0 - (1 - 1) = 0$. The denominator becomes $\alpha(0)(e^{3(0)} - 1) = 0(1 - 1) = 0$. Thus, the limit is of the indeterminate form $\frac{0}{0}$. This indicates that we need to simplify the expression or use L'Hôpital's Rule or Taylor series expansion.

Step 2: Rewrite the expression to utilize standard limits. We can split the fraction into two parts: $\beta = \mathop {\lim }\limits_{x \to 0} \left( \frac{\alpha x}{\alpha x(e^{3x}-1)} - \frac{e^{3x}-1}{\alpha x(e^{3x}-1)} \right)$ $\beta = \mathop {\lim }\limits_{x \to 0} \left( \frac{1}{e^{3x}-1} - \frac{1}{\alpha x} \right)$ This form still appears problematic as $x \to 0$. Let's try a different manipulation by dividing the numerator and denominator by $x$.

Step 3: Manipulate the expression to use the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{e^y - 1}{y} = 1$. Divide the numerator and the denominator by $x$: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{\frac{\alpha x - (e^{3x}-1)}{x}}{\frac{\alpha x(e^{3x}-1)}{x}}$ $\beta = \mathop {\lim }\limits_{x \to 0} \frac{\alpha - \frac{e^{3x}-1}{x}}{\alpha (e^{3x}-1)}$ As $x \to 0$, the denominator $\alpha (e^{3x}-1) \to \alpha(0) = 0$. For the limit $\beta$ to exist and be finite, the numerator must also approach 0. Let's examine the numerator: $\alpha - \mathop {\lim }\limits_{x \to 0} \frac{e^{3x}-1}{x}$. We know that $\mathop {\lim }\limits_{x \to 0} \frac{e^{3x}-1}{x} = \mathop {\lim }\limits_{x \to 0} 3 \cdot \frac{e^{3x}-1}{3x}$. Let $y = 3x$. As $x \to 0$, $y \to 0$. So, $\mathop {\lim }\limits_{x \to 0} 3 \cdot \frac{e^{3x}-1}{3x} = 3 \cdot \mathop {\lim }\limits_{y \to 0} \frac{e^y-1}{y} = 3 \cdot 1 = 3$. For the numerator to be 0, we must have $\alpha - 3 = 0$, which implies $\alpha = 3$.

Step 4: Evaluate the limit with $\alpha = 3$. Now substitute $\alpha = 3$ back into the limit expression derived in Step 3: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{3 - \frac{e^{3x}-1}{x}}{3 (e^{3x}-1)}$ As $x \to 0$, the numerator is $3 - 3 = 0$, and the denominator is $3(1-1) = 0$. We still have the $\frac{0}{0}$ indeterminate form. This suggests we need to use the Taylor series expansion or L'Hôpital's Rule.

Step 5: Use Taylor Series Expansion for $e^{3x}$. The Taylor series expansion of $e^y$ around $y=0$ is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$. Substituting $y = 3x$, we get: $e^{3x} = 1 + (3x) + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + \dots$ $e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \dots$ Therefore, $e^{3x} - 1 = 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \dots$.

Step 6: Substitute the Taylor expansion into the original limit expression. Let's go back to the original expression and substitute the expansion for $e^{3x}-1$: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{\alpha x - \left(3x + \frac{9x^2}{2} + \dots \right)}{\alpha x \left(3x + \frac{9x^2}{2} + \dots \right)}$ We already determined that for the limit to exist, $\alpha$ must be 3. Substitute $\alpha=3$: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{3x - \left(3x + \frac{9x^2}{2} + \dots \right)}{3x \left(3x + \frac{9x^2}{2} + \dots \right)}$ $\beta = \mathop {\lim }\limits_{x \to 0} \frac{-\frac{9x^2}{2} - \dots}{9x^2 + \frac{27x^3}{2} + \dots}$ Now, divide the numerator and the denominator by the lowest power of $x$ in the denominator, which is $x^2$: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{-\frac{9}{2} - \dots}{9 + \frac{27x}{2} + \dots}$ As $x \to 0$, all terms with $x$ will go to zero. $\beta = \frac{-\frac{9}{2}}{9} = -\frac{9}{2 \cdot 9} = -\frac{1}{2}$

Step 7: Calculate $\alpha + \beta$. We found $\alpha = 3$ and $\beta = -\frac{1}{2}$. Therefore, $\alpha + \beta = 3 + \left(-\frac{1}{2}\right) = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2}$.

Alternative Approach using L'Hôpital's Rule:

Step 1 (Alternative): Verify indeterminate form and apply L'Hôpital's Rule. As shown before, the limit is of the form $\frac{0}{0}$. We can apply L'Hôpital's Rule. Let $f(x) = \alpha x - (e^{3x}-1)$ and $g(x) = \alpha x (e^{3x}-1) = \alpha x e^{3x} - \alpha x$. $f'(x) = \alpha - 3e^{3x}$. $g'(x) = \alpha (1 \cdot e^{3x} + x \cdot 3e^{3x}) - \alpha = \alpha e^{3x} + 3\alpha x e^{3x} - \alpha$.

$\beta = \mathop {\lim }\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to 0} \frac{\alpha - 3e^{3x}}{\alpha e^{3x} + 3\alpha x e^{3x} - \alpha}$ Substitute $x=0$: Numerator: $\alpha - 3e^0 = \alpha - 3$. Denominator: $\alpha e^0 + 3\alpha (0) e^0 - \alpha = \alpha + 0 - \alpha = 0$. For the limit to exist, the numerator must also be 0. So, $\alpha - 3 = 0 \implies \alpha = 3$.

Step 2 (Alternative): Apply L'Hôpital's Rule again with $\alpha = 3$. With $\alpha = 3$, the expression becomes: $\beta = \mathop {\lim }\limits_{x \to 0} \frac{3 - 3e^{3x}}{3e^{3x} + 9x e^{3x} - 3}$ This is still of the form $\frac{0}{0}$. Apply L'Hôpital's Rule again. $f''(x) = \frac{d}{dx}(3 - 3e^{3x}) = -9e^{3x}$. $g''(x) = \frac{d}{dx}(3e^{3x} + 9x e^{3x} - 3) = 9e^{3x} + 9(1 \cdot e^{3x} + x \cdot 3e^{3x}) = 9e^{3x} + 9e^{3x} + 27x e^{3x} = 18e^{3x} + 27x e^{3x}$.

$\beta = \mathop {\lim }\limits_{x \to 0} \frac{-9e^{3x}}{18e^{3x} + 27x e^{3x}}$ Substitute $x=0$: $\beta = \frac{-9e^0}{18e^0 + 27(0)e^0} = \frac{-9}{18 + 0} = \frac{-9}{18} = -\frac{1}{2}$

Step 3 (Alternative): Calculate $\alpha + \beta$. We found $\alpha = 3$ and $\beta = -\frac{1}{2}$. Therefore, $\alpha + \beta = 3 + \left(-\frac{1}{2}\right) = \frac{5}{2}$.

Common Mistakes & Tips

• Incorrectly handling indeterminate forms: Directly substituting $x=0$ can lead to incorrect conclusions. Always identify the indeterminate form ($\frac{0}{0}, \frac{\infty}{\infty}$, etc.) and apply appropriate techniques.
• Errors in Taylor series expansion: Ensure the Taylor series for $e^y$ is correctly recalled and applied, especially the coefficients and powers of $y$. For $e^{3x}$, the expansion starts with $1 + 3x + \frac{(3x)^2}{2!} + \dots$.
• Algebraic errors: Be meticulous with algebraic manipulations, especially when dividing by terms involving $x$ or simplifying fractions.
• L'Hôpital's Rule application: Remember to differentiate the numerator and denominator separately. If the form remains indeterminate after the first application, apply it again.

Summary

The problem requires evaluating a limit of an indeterminate form. We first analyzed the limit expression and determined that for the limit $\beta$ to be finite, the parameter $\alpha$ must be equal to 3. This was achieved by ensuring the numerator of the limit expression approached zero when the denominator approached zero. Once $\alpha$ was found, we evaluated the limit $\beta$ using either the Taylor series expansion of $e^{3x}$ or by applying L'Hôpital's Rule twice. Both methods yielded $\beta = -\frac{1}{2}$. Finally, we calculated the sum $\alpha + \beta = 3 + (-\frac{1}{2}) = \frac{5}{2}$.

The final answer is $\boxed{\frac{5}{2}}$.