For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$

The function $f$ is NOT differentiable at $x = 0$

There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$

For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$

$x = 0$ is a point of local minima of $f$

(P) → (1)   (Q) → (3)   (R) → (2)   (S) → (5)

(P) → (2)   (Q) → (1)   (R) → (4)   (S) → (3)

(P) → (5)   (Q) → (1)   (R) → (4)   (S) → (3)

(P) → (2)   (Q) → (3)   (R) → (1)   (S) → (5)