Calculate $\:\lim_\limits{ \Large_{z \to c}}\:\:{\frac{z^n - c^n}{z-c}} ~~ c,z \in \mathbb{C}\:$ without L'Hopital's rule

Question

Without any concept of differentiation, I need to evaluate the value of the following limit:

$$\lim_{ \Large_{z \to c}}\:\:{\frac{z^n - c^n}{z - c}} ~~~~~~~~ c,z \in \mathbb{C}$$

I tried to transform the term but got to nothing useful out of it. Can someone give me a hint/solution on how to tackle this problem?

Answer 1

One has $$(z^n - c^n) = (z-c)(z^{n-1} + z^{n-2}c + \ldots + zc^{n-2} + c^{n-1})$$ which is proven by multiplying out the right hand side and observing a cascade of cancellations.

Answer 2

For $c\ne0$,$$\lim_{z \to c}{\frac{z^n + c^n}{z+c}}=\\ \frac{\lim_{z \to c}{(z^n + c^n)}}{\lim_{z \to c}{(z+c)}}=\\ \frac{\lim_{z \to c}{z^n + \lim_{z \to c}c^n}}{\lim_{z \to c}{z+\lim_{z \to c}c}}=\\ \frac{c^n+c^n}{c+c}=\\ c^{n-1}.$$

For $c=0$ and $n>0$,

$$\lim_{z \to 0}{\frac{z^n}{z}}=\lim_{z \to 0}z^{n-1}=0.$$

For $c=0$ and $n=0$,

$$\lim_{z \to 0}{\frac1z}$$ isn't defined.

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Now for the "real" question, use the change of variable $z=c+w$, and by the Binomial theorem

$$\lim_{z\to c}\frac{z^n-c^n}{z-c}=\\ \lim_{w\to0}\frac{(c+w)^n-c^n}w=\\ \lim_{w\to0}\frac{c^n+nwc^{n-1}+\frac{n(n-1)}2w^2c^{n-2}+\cdots w^n-c^n}w =\\ \lim_{w\to0}\left(nc^{n-1}+\frac{n(n-1)}2wc^{n-2}+\frac{n(n-1)(n-2)}{3!}w^2c^{n-3}\cdots+w^{n-1}\right)=\\nc^{n-1}+0+0+\cdots0.$$

For $c=0$ the reasoning is as above.

Answer 3

If you did not miswrite the equation, you can just set z = c, and get 2c^n/2c = c^(n-1).

L'hopital cannot even be used, as it only is working if the term approaches 0/0, which it doesn't.

Answer 4

Factorization: $\frac{a^{n}-b^{n}}{a-b}=(a^{n-1}+ba^{n-2}+b^{2}a^{n-3}+\ldots +b^{n-2}a+b^{n-1})$

So for your limit: $\lim_{z\to c}{\frac{z^{n}-c^{n}}{z-c}}=(n)*c^{n-1}$