Solve the limit problems.

Steps to Solve

1. Evaluate $\lim_{x\to 0} \frac{x}{|x| + x^2}$

   We need to consider the limit from the left and right separately because of the absolute value.

   • For $x \to 0^+$, $|x| = x$, so the expression becomes $\frac{x}{x + x^2} = \frac{x}{x(1+x)} = \frac{1}{1+x}$. Then, (\lim_{x\to 0^+} \frac{1}{1+x} = \frac{1}{1+0} = 1).

   • For $x \to 0^-$, $|x| = -x$, so the expression becomes $\frac{x}{-x + x^2} = \frac{x}{x(-1+x)} = \frac{1}{-1+x}$. Then, (\lim_{x\to 0^-} \frac{1}{-1+x} = \frac{1}{-1+0} = -1). Since the limits from the left and right are different, the limit does not exist.

2. Evaluate $\lim_{x\to 0} \frac{\sin(\pi \cos^2 x)}{x^2}$

   Rewrite $\cos^2 x$ as $1 - \sin^2 x$. So, $\sin(\pi \cos^2 x) = \sin(\pi (1 - \sin^2 x)) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x)$.

   Using the small angle approximation $\sin(x) \approx x$ for $x \to 0$, we can approximate $\sin(\pi \sin^2 x)$ as $\pi \sin^2 x$. Then, $\lim_{x\to 0} \frac{\sin(\pi \cos^2 x)}{x^2} = \lim_{x\to 0} \frac{\pi \sin^2 x}{x^2} = \pi \lim_{x\to 0} \frac{\sin^2 x}{x^2} = \pi \left( \lim_{x\to 0} \frac{\sin x}{x} \right)^2$.

   Since $\lim_{x\to 0} \frac{\sin x}{x} = 1$, we have $\pi (1)^2 = \pi$.

3. Find $\lim_{a\to 0^+} \alpha(a)$ and $\lim_{a\to 0^+} \beta(a)$

   Let $A = \sqrt[6]{1+a} - 1$ and $B = \sqrt{1+a} - 1$. The given quadratic equation can be written as $Ax^2 + Bx + A = 0$.

   Using the quadratic formula, the roots are given by $x = \frac{-B \pm \sqrt{B^2 - 4A^2}}{2A}$. $x = \frac{-(\sqrt{1+a}-1) \pm \sqrt{(\sqrt{1+a}-1)^2 - 4(\sqrt[6]{1+a}-1)^2}}{2(\sqrt[6]{1+a}-1)}$.

   Let's use the binomial approximation: $\sqrt[n]{1+a} \approx 1 + \frac{a}{n}$ for small $a$.

   Then, $A = \sqrt[6]{1+a} - 1 \approx 1 + \frac{a}{6} - 1 = \frac{a}{6}$ and $B = \sqrt{1+a} - 1 \approx 1 + \frac{a}{2} - 1 = \frac{a}{2}$.

   Substitute these approximations into the roots equation: $x = \frac{-\frac{a}{2} \pm \sqrt{(\frac{a}{2})^2 - 4(\frac{a}{6})^2}}{2(\frac{a}{6})} = \frac{-\frac{a}{2} \pm \sqrt{\frac{a^2}{4} - \frac{4a^2}{36}}}{\frac{a}{3}} = \frac{-\frac{a}{2} \pm \sqrt{\frac{9a^2 - 4a^2}{36}}}{\frac{a}{3}} = \frac{-\frac{a}{2} \pm \sqrt{\frac{5a^2}{36}}}{\frac{a}{3}}$

   $x = \frac{-\frac{a}{2} \pm \frac{a\sqrt{5}}{6}}{\frac{a}{3}} = \frac{a(-\frac{1}{2} \pm \frac{\sqrt{5}}{6})}{\frac{a}{3}} = 3(-\frac{1}{2} \pm \frac{\sqrt{5}}{6}) = -\frac{3}{2} \pm \frac{\sqrt{5}}{2}$.

   The two roots are $\alpha(a) = -\frac{3}{2} + \frac{\sqrt{5}}{2}$ and $\beta(a) = -\frac{3}{2} - \frac{\sqrt{5}}{2}$. Since these are constants, the limits are: $\lim_{a\to 0^+} \alpha(a) = -\frac{3}{2} + \frac{\sqrt{5}}{2} \approx -0.38$ $\lim_{a\to 0^+} \beta(a) = -\frac{3}{2} - \frac{\sqrt{5}}{2} \approx -2.62$

   However, the available answers do not match these values and the binomial approximation might not be precise enough. Let's try substituting $a=0$ to $Ax^2 + Bx + A = 0$. $A = \sqrt[6]{1+0}-1 = 0$ and $B = \sqrt{1+0}-1=0$ so we can't do this.

   Instead, divide the original equation by $A$: $x^2 + \frac{B}{A}x + 1 = 0$ where $\frac{B}{A} = \frac{\sqrt{1+a}-1}{\sqrt[6]{1+a}-1} = \frac{(\sqrt{1+a}-1)(\sqrt{1+a}+1)(\sqrt[6]{1+a}+1)}{(\sqrt[6]{1+a}-1)(\sqrt{1+a}+1)(\sqrt[6]{1+a}+1)} = \frac{a}{(\sqrt[6]{1+a}-1)(\sqrt{1+a}+1)}$.

   Multiply and divide by $(\sqrt[6]{1+a}^5+\sqrt[6]{1+a}^4+\sqrt[6]{1+a}^3+\sqrt[6]{1+a}^2+\sqrt[6]{1+a}+1)$ to get $\frac{B}{A} = \frac{a}{\frac{a}{6} (\sqrt{1+a}+1)}\approx\frac{6}{(\sqrt{1+a}+1)}$.

   As a approaches zero, we have $\frac{B}{A} \to \frac{6}{2} = 3$.

   Thus $x^2 + 3x + 1 = 0$. Using the quadratic formula we get: $x = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}$. Therefore the limits are $\frac{-3 + \sqrt{5}}{2}$ and $\frac{-3 - \sqrt{5}}{2}$.