Limit of $\frac{1}{x}(1-\alpha^{x})$

Question

Consider the following function $f$ for a given constant $0<\alpha <1$:

$f:(0,1) \to \mathbb{R}$, where $f(x)=\frac{1}{x}(1-\alpha^{x})$.

Now we plug in a point $x_0$ and then take $\lim_{\alpha \to 0}f(x_0)=\frac{1}{x_0}(1-\alpha^{x_0})$.

My intuition is that the limit must be $\frac{1}{x_0}$ because the expression $\alpha^{x_0}$ goes to $0$. However, in our lecture the professor included the follwoing the step:

$\lim_{\alpha \to 0}f(x_0)=\frac{1}{x_0}(1-\alpha^{x_0})=\frac{1}{x_0}(1-e^{x_0 \ln(\alpha)})$.

Then he argued that because of $\lim_{\alpha \to 0}\ln(\alpha)=-\infty$ and $\lim_{t \to -\infty}e^t=0$ we can conclude that $\lim_{\alpha \to 0}f(x_0)=\frac{1}{x_0}(1-\alpha^{x_0})=\frac{1}{x_0}$.

Why do we need this additional step?

Answer 1

What you observe is correct. Your professor overcomplicated the case, given that you already know that

$$\lim_{\alpha \to 0} \alpha^x = 0$$

where $x > 0$.