If $$\alpha = \lim_{x \to 0^+} \left(\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}}\right)$$ and $$\beta = \lim_{x \to 0} (1 + \sin x)^{\frac{1}{2}\cot x}$$ are the roots of the quadratic equation $$ax^2 + bx - \sqrt{e} = 0$$, then $$12\log_e(a + b)$$ is equal to _____

We have two limits to evaluate.

$$\alpha = \lim_{x \to 0^{+}}\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}$$

Write $$f(u)=e^{u}$$. Near $$x=0$$, both $$u_1=\sqrt{\tan x}$$ and $$u_2=\sqrt{x}$$ approach $$0$$. For a differentiable function, $$\displaystyle\lim_{u_1,u_2\to 0}\frac{f(u_1)-f(u_2)}{u_1-u_2}=f'(0)$$, because the expression is the symmetric derivative at the point $$0$$.

Here $$f'(u)=e^{u}$$, so $$f'(0)=1$$. Hence

$$\alpha = 1$$.

For the second limit

$$\beta = \lim_{x \to 0}\left(1+\sin x\right)^{\frac12\cot x}$$

take natural logarithm: $$\ln\beta = \lim_{x\to 0}\left(\frac12\cot x\right)\ln(1+\sin x)$$.

Series expansions about $$x=0$$:

$$\sin x = x-\frac{x^3}{6}+O(x^5)$$, $$\cos x = 1-\frac{x^2}{2}+O(x^4)$$, $$\cot x = \frac{\cos x}{\sin x} = \frac1x-\frac{x}{3}+O(x^3)$$.

Thus $$\frac12\cot x=\frac1{2x}-\frac{x}{6}+O(x^3)$$.

Also $$\ln(1+\sin x)=\sin x-\frac{\sin^2 x}{2}+O(x^3) = x-\frac{x^2}{2}-\frac{x^3}{6}+O(x^3).$$

Multiply the two series:

$$\Bigl(\frac1{2x}-\frac{x}{6}+O(x^3)\Bigr) \Bigl(x-\frac{x^2}{2}+O(x^3)\Bigr) = \frac12 + O(x).$$

Therefore $$\ln\beta \to \frac12$$ and

$$\beta = e^{1/2} = \sqrt{e}.$$

The numbers $$\alpha=1$$ and $$\beta=\sqrt{e}$$ are roots of the quadratic

$$ax^2 + bx - \sqrt{e}=0.$$

Using the relations for roots of $$Ax^2+Bx+C=0$$:
Sum of roots $$= -\frac{B}{A}$$, product of roots $$=\frac{C}{A}$$.

Sum: $$1+\sqrt{e} = -\frac{b}{a} \quad -(1)$$
Product: $$1\cdot\sqrt{e} = \frac{-\sqrt{e}}{a} \quad -(2)$$.

From $$(2)$$, $$a = -1.$$ Substituting in $$(1)$$ gives $$b = 1+\sqrt{e}.$$

Hence $$a+b = -1 + 1+\sqrt{e} = \sqrt{e}.$$

Finally,

$$12\log_e(a+b) = 12\log_e\!\bigl(\sqrt{e}\bigr) = 12\cdot\frac12 = 6.$$

So the required value is $$6$$.