Key Concepts and Formulas

• Limit Definition: The limit of a function as $x$ approaches a certain value describes the behavior of the function near that value.
• Standard Limits:
  • $\lim_{y \to 0} \frac{e^y - 1}{y} = 1$
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \cos x = 1$
• Logarithm Properties: $\log_b a^c = c \log_b a$.
• Quadratic Equation Properties: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha \beta = \frac{c}{a}$.
• Exponential Form of Limits: $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) \ln(f(x))}$, especially useful for indeterminate forms of type $1^\infty$.

Step-by-Step Solution

Step 1: Evaluate the limit for $\alpha$. We are given \alpha=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right). This is of the indeterminate form $\frac{e^0 - e^0}{0-0} = \frac{0}{0}$. We can rewrite the expression by factoring out $e^{\sqrt{x}}$ from the numerator: $\alpha = \lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{\sqrt{\tan x}-\sqrt{x}}$ Let $y = \sqrt{\tan x} - \sqrt{x}$. As $x \rightarrow 0^{+}$, we have $\tan x \rightarrow 0$ and $\sqrt{\tan x} \rightarrow 0$, and $\sqrt{x} \rightarrow 0$. To evaluate the behavior of $y$ as $x \rightarrow 0^{+}$, we can use Taylor series expansions for $\tan x$ around $x=0$. $\tan x = x + \frac{x^3}{3} + O(x^5)$ $\sqrt{\tan x} = \sqrt{x + \frac{x^3}{3} + O(x^5)} = \sqrt{x}\sqrt{1 + \frac{x^2}{3} + O(x^4)}$ Using the binomial approximation $(1+u)^n \approx 1 + nu$ for small $u$: $\sqrt{\tan x} \approx \sqrt{x} \left(1 + \frac{1}{2} \cdot \frac{x^2}{3}\right) = \sqrt{x} \left(1 + \frac{x^2}{6}\right) = \sqrt{x} + \frac{x^{2.5}}{6}$ So, $y = \sqrt{\tan x} - \sqrt{x} = (\sqrt{x} + \frac{x^{2.5}}{6}) - \sqrt{x} = \frac{x^{2.5}}{6}$. As $x \rightarrow 0^{+}$, $y \rightarrow 0$. Now, consider the limit: $\alpha = \lim _{x \rightarrow 0^{+}} e^{\sqrt{x}} \cdot \lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}-\sqrt{x}}-1}{\sqrt{\tan x}-\sqrt{x}}$ The first limit is $e^{\sqrt{0}} = e^0 = 1$. The second limit is of the form $\lim_{y \to 0} \frac{e^y - 1}{y}$, which is 1. Therefore, $\alpha = 1 \cdot 1 = 1$.

Step 2: Evaluate the limit for $\beta$. We are given \beta=\lim _\limits{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}. This is of the indeterminate form $(1+0)^{\frac{1}{2} \cdot \infty} = 1^\infty$. We use the property $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) \ln(f(x))}$. Here, $f(x) = 1 + \sin x$ and $g(x) = \frac{1}{2} \cot x$. $\beta = e^{\lim _{x \rightarrow 0} \left(\frac{1}{2} \cot x\right) \ln(1+\sin x)}$ Let's evaluate the exponent: $L = \lim _{x \rightarrow 0} \frac{1}{2} \cot x \ln(1+\sin x) = \lim _{x \rightarrow 0} \frac{1}{2} \frac{\cos x}{\sin x} \ln(1+\sin x)$ $L = \frac{1}{2} \lim _{x \rightarrow 0} \cos x \cdot \lim _{x \rightarrow 0} \frac{\ln(1+\sin x)}{\sin x}$ The first limit is $\cos 0 = 1$. For the second limit, let $u = \sin x$. As $x \rightarrow 0$, $u \rightarrow 0$. The limit becomes $\lim_{u \to 0} \frac{\ln(1+u)}{u}$. This is a standard limit, which is equal to 1. So, $L = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$. Therefore, $\beta = e^{1/2} = \sqrt{e}$.

Step 3: Use the properties of the quadratic equation. We are given that $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx - \sqrt{e} = 0$. From the equation, the constant term is $c = -\sqrt{e}$. The product of the roots is $\alpha \beta = \frac{c}{a}$. We have $\alpha = 1$ and $\beta = \sqrt{e}$. So, $(1)(\sqrt{e}) = \frac{-\sqrt{e}}{a}$. $\sqrt{e} = \frac{-\sqrt{e}}{a}$ Since $\sqrt{e} \neq 0$, we can divide both sides by $\sqrt{e}$: $1 = \frac{-1}{a} \implies a = -1$.

The sum of the roots is $\alpha + \beta = -\frac{b}{a}$. We have $\alpha = 1$ and $\beta = \sqrt{e}$. So, $1 + \sqrt{e} = -\frac{b}{a}$. Substitute the value of $a = -1$: $1 + \sqrt{e} = -\frac{b}{-1}$ $1 + \sqrt{e} = b$.

Step 4: Calculate the final expression. We need to find $12 \log_{e}(a+b)$. We found $a = -1$ and $b = 1 + \sqrt{e}$. So, $a+b = -1 + (1+\sqrt{e}) = \sqrt{e}$. Now, calculate $12 \log_{e}(a+b)$: $12 \log_{e}(\sqrt{e})$ Using the property $\log_b a^c = c \log_b a$, we have $\log_{e}(\sqrt{e}) = \log_{e}(e^{1/2}) = \frac{1}{2} \log_{e}(e)$. Since $\log_{e}(e) = 1$, we get $\log_{e}(\sqrt{e}) = \frac{1}{2}$. Therefore, the expression is $12 \cdot \frac{1}{2} = 6$.

Common Mistakes & Tips

• Incorrectly applying standard limits: Ensure that the arguments of the standard limits match the form of the expression. For instance, in the limit for $\alpha$, it's crucial to recognize that $\sqrt{\tan x} - \sqrt{x}$ approaches 0 as $x \rightarrow 0^{+}$.
• Algebraic errors in solving for $a$ and $b$: Double-check the signs and substitutions when using the sum and product of roots formulas.
• Simplifying logarithmic terms: Remember that $\log_e e = 1$ and $\log_e e^k = k$.

Summary

The problem involves evaluating two limits to find the roots of a quadratic equation. The first limit, $\alpha$, is found by algebraic manipulation and applying the standard limit $\lim_{y \to 0} \frac{e^y - 1}{y} = 1$. The second limit, $\beta$, is evaluated using the property for limits of the form $1^\infty$, transforming it into an exponential form and using the standard limit $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$. Once $\alpha$ and $\beta$ are determined, they are used as roots of the given quadratic equation $ax^2 + bx - \sqrt{e} = 0$. By applying the sum and product of roots formulas, the coefficients $a$ and $b$ are found. Finally, the expression $12 \log_{e}(a+b)$ is computed using the values of $a$ and $b$.

The final answer is $\boxed{6}$.