Solving limit for a finite value

Question

If $a,b$ belong to $\{-1,0,1,2\}$ then for how many ordered pairs $(a,b)$ does $\lim_{x \to 0} f(x) = \frac{e^{ax} -1}{x^b}$ have a finite value?

Answer 1

Notice that the limit is $$\lim_{x \to 0} \frac{e^{a x} - e^{a \times 0}}{x} \times \frac{1}{x^{b-1}}$$

Now, the first term is exactly the thing which, when we send $x \to 0$, we obtain $\dfrac{d}{dx} e^{ax}$ evaluated at $x=0$; i.e. $a$.

As long as $a$ is not $0$, then, the limit of the product will be the product of the limits: $$a \times \lim_{x \to 0} x^{1-b}$$ whether that be $a \times 0$ (if $b < 1$), $a \times 1$ (if $b = 1$) or $a \times \infty$ (if $b > 1$).

The only problem is therefore if $a = 0$, when $f(x) = 0$ and so the limit is $0$.

Answer 2

If $a=0$, there is nothing to do.

Assume now that $a\neq0$ ;when $x$ belongs to a neighborhood of zero, we can use the Taylor expansion of exponential $$\mathrm{e}^{ax}=1+ax+\frac{a^2x^2}{2}+O(x^3)$$ so that \begin{align} \frac{\mathrm{e}^{ax}-1}{x^b} &=\frac{ax+\frac{a^2x^2}{2}+O(x^3)}{x^b} \\ &=ax^{1-b}+a^2\frac{x^{2-b}}{2}+O\left(x^{3-b}\right). \end{align} Thus, the limit exists iff $b\leq1$.