Final Value Theorem Z Transform This is probably really easy but I cannot see the wood for the trees.
I see the discrete time final value theorem given as:
$$ \lim_{k\to \infty}f[k] = \lim_{z\to 1} (z-1)F(z)$$
BUT also expressed as,
$$ \lim_{k\to \infty}f[k] = \lim_{z\to 1} (1-z^{-1})F(z)$$
But, 
$$\frac{1}{1} - \frac{1}{z} = \frac{z-1}{z}$$
Not $z - 1$ 
Therefore they are not equivalent. Am I missing something here!?
I feel stupid asking the question. 
Thanks for your help
 A: As $z \rightarrow 1$
$$\dfrac{z-1}{z} \rightarrow \dfrac{z-1}{1} = z-1$$
Update: Adding a proof of the final value theorem
Start with the definition of the z-transform
$$F(z) = \lim_{N \to \infty} \sum_{k=0}^{N} f[k]z^{-k}$$
and also use the time advance theorem of the z-transform
$$z\left[F(z)-f[0]\right] = \lim_{N \to \infty} \sum_{k=0}^{N} f[k+1]z^{-k}$$
Now subtract the first from the second
$$(z-1)F(z) - zf[0] = \lim_{N \to \infty} \sum_{k=0}^{N} f[k+1]z^{-k}-f[k]z^{-k}$$
Then take the limit as $z \to 1$
$$\lim_{z \to 1} \space (z-1)F(z) - f[0] = \lim_{N \to \infty} \space f[N+1]-f[0]$$
so
$$\lim_{z \to 1} \space (z-1)F(z) = \lim_{N \to \infty} \space f[N+1]=\lim_{N \to \infty} \space f[N]$$
Alternately
Using the time delay theorem of the z-transform
$$z^{-1}F(z) = \lim_{N \to \infty} \sum_{k=0}^{N} f[k-1]z^{-k}$$
we can get to
$$(1-z^{-1})F(z) = \lim_{N \to \infty} \sum_{k=0}^{N} f[k]z^{-k} -f[k-1]z^{-k}$$
and assuming $f[k] = 0$ for $k <0$ and taking the limit as $z\to1$
$$\lim_{z \to 1} \space (1-z^{-1})F(z) = \lim_{N \to \infty} \space f[N]-f[-1] = \lim_{N \to \infty} \space f[N]$$
