limit calculation using l'Hopital rule

Question

A generalized logarithmic function looks like this:

$$\frac{(1-h)^{1-x}}{1-x}$$

Apparently if you take the limit of $x$ to $1$, the result is $\ln(1-h)$. However, I do not get this result. Somewhere in between u get 0/0, so you apply l'Hopital, and you take the derivative of the nominator and denominator separately, but then I go wrong somewhere...

Answer 1

Recall that

$$(1-h)^{1-x}=e^{(1-x)\log (1-h)}\implies [(1-h)^{1-x}]'=[e^{(1-x)\log (1-h)}]'=(1-h)^{1-x}\log(1-h)$$

and then by l'Hopital

$$\lim_{x\to 1}\frac{(1-h)^{1-x}-1}{ 1-x}=\lim_{x\to 1}\frac{(1-h)^{1-x}\log(1-h)}{ -1}=-\log (1-h)$$

Answer 2

We assume $0<h<1$, one may write $$ (1-h)^{1-x}=e^{(1-x)\ln (1-h)} $$ giving, by differentiating with respect to $x$ $$ \left((1-h)^{1-x} \right)'=\left(e^{(1-x)\ln (1-h)}\right)'=\ln (1-h)\cdot e^{(1-x)\ln (1-h)} $$ then by using L'Hospital's rule $$ \lim_{x \to 1}\frac{(1-h)^{1-x}}{1-x}=\lim_{x \to 1}\frac{\ln (1-h)\cdot(1-h)^{1-x}}{-1}=\cdots. $$Hope you can finish it.