Evaluating $\lim_{\alpha\to 1}\frac{x_0^{1-\alpha}x^{\alpha}-x}{1-\alpha}$ I have an engineering equation:
$$y=\frac{x_0^{1-\alpha}x^{\alpha}-x}{1-\alpha}$$
where $x_0$ is a constant and $\alpha$ is a parameter.
A potential singularity arises for $\alpha =1$ because then:
$$y=\frac{x-x}{1-1}=\frac{0}{0}$$
Can anyone show me how to determine:
$$\lim_{\alpha \to 1}y$$
Thank you.
 A: You have already shown that you have an indeterminate form
Applying L'Hopital's rule:
The derivative of the numerator $\frac {d}{da} e^{(1-a)\ln x_0+a\ln x} - x = x_0^{1-a}x^a(\ln x-\ln x_0)$
as $a$ approaches $1,$ we get: $x(\ln x - \ln x_0)$
and the derivative of the denominator is $-1$
$x(\ln x_0 - \ln x)$
A: Note that $x=x^ax^{1-a}$
Thus $$\frac{x_0^{1-\alpha}x^{\alpha}-x}{1-\alpha}=\frac{x^{\alpha}(x_0^{1-\alpha}-x^{1-\alpha})}{1-\alpha}=\frac{x^{1-t}(x_0^{t}-x^{t})}{t}=\frac{x^{1-t}((\frac{x_0}{x})^t-1)}{t}$$
where $t=1-a$ and $t \to 0$ as $a \to 1$

Thus $$\lim_{t \to 0}\frac{x^{1-t}((\frac{x_0}{x})^t-1)}{t}=xf'(0)$$ where $f(t)=(\frac{x_0}{x})^t$

A: For the limit itself, you already received good and simple answers.
Since this is an engineering problem, you could be interested by the fact that Taylor expansions built at $\alpha=1$ can give you the limit but also how it is approached.
$$y=x \log \left(\frac{x_0}{x}\right)\left(1-\frac{1}{2} \log \left(\frac{x_0}{x}\right)(\alpha -1)+O\left((\alpha -1)^2\right)\right)$$
A: The notation is quite confusing and I think it is simpler if you write the problem as
$f(x) = \frac{ac^x-b}{1-x}$, where $a=x_0, b=x, c=\frac{x}{x_0}.$
$$\lim_{x\to 1} f(x)$$ now looks elementary(doesn't it?).
