Limit calculation (de l’Hospital?) 
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$$\lim_{x \rightarrow 0} \frac{{(1-x)}^{k}-({1-k x})}{x^2}$$ $( k>0)$
2)
$$\lim_{x \rightarrow z} \frac{x^m-z^m}{x^n-z^n}$$ $(z>0, m, n \in \mathbb{R}, n \neq 0)$

What I’ve done for: for 1)
I have applied l’Hospital, then I have
$\lim_{x  \rightarrow 0} \frac{k (1-x)^{k-1} (-1)+k}{2x} = \lim_{x  \rightarrow 0} \frac{k ((x-1)^{k-1}+1)}{2x}$ but I don’t know how to continue from here. If $x$ converges to zero the nominator also converges to zero, right?
And for the second one, I have also applied l’Hospital
$$\lim_{x \rightarrow z} \frac{x^m-z^m}{x^n-z^n}= \lim_{x \rightarrow z} \frac{mx^{m-1} - mz^{m-1}}{nx^{n-1}-nz^{n-1}}$$ but it doesn’t look like that this would help me?
 A: Lemma: $\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$
Proof: Use the fact that $x^n-a^n = (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\dots+a^{n-2}x+a^{n-1})$, if $n$ is a natural number. Otherwise use L'Hopital rule.
Now our first limit after your use of L'Hopital rule becomes
\begin{align}
\lim_{x \rightarrow 0} \frac{{(1-x)}^{k}-({1-k x})}{x^2} &= \frac k2\lim_{1-x\to1}\frac{1-(1-x)^{k-1}}{1-(1-x)}\\
&=\frac k2\cdot (k-1)1^{k-2} \\
&= \frac{k(k-1)}{2}
\end{align}
Second limit immediately becomes
$$\lim_{x\to z}\dfrac{\dfrac{x^m-z^m}{x-z}}{\dfrac{x^n-z^n}{x-z}}=\frac mnz^{m-n}$$
A: Your assertion that if denominator tends to $0$, so should numerator, is in general, baseless, although under certain conditions it holds. Moreover, you have incorrectly assumed that $(1-x)^k(-1)=(x-1)^k$. This holds only if:
$$(1-x)^k(-1)=(x-1)^k(-1)^{k+1}=(x-1)^k$$
Which means that $k$ is odd. However, if you proceed with $(1-x)^k(-1)$, you will see that $\frac 00$ indeterminate form is retained and hence L'Hôpital rule can be applied again.
Alternative:
Instead of applying the L'Hôpital rule, it would be more useful to take advantage of Newton's expansion:
$$(1-x)^k=1-kx+\frac {k(k-1) x^2}{2}-...$$
This is an infinite series, but writing three terms suffices for our purpose.
For the second one, you have not correctly applied the rule. The differentiation is to be done with respect to $x$, and $z$ is to be simply treated as a constant. I believe you can do it now, the value of the limit would depend upon which of $m$ and $n$ is greater.
