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.
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 $-1$

$x(\ln x_0 - \ln x)$
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?).