$\lim_{x \to 1} \frac{x^k - 1}{x - 1}$ using $\delta, \epsilon$ Hi all, 
 I think I have a pretty good start to the problem I listed above, but wanted some help with manipulating the end. I believe (from a hint) that I can somehow use the Difference Power Formula to simplify it more.
Proof:
 We look at the point $x_0 = 0$, and look at a potential limit $L=k$.
Suppose $\epsilon > 0$ and $\delta = $____
We must show that for all $|x-x_0| < \delta$, it holds that $|f(x) - L| < \epsilon$.
Suppose $|x-x_0| \to |x-1| < \delta$. 
Then $|f(x) - L| \to |\frac{x^k-1}{x-1}-k| \to |\frac{x^k-1-k(x-1)}{x-1}|$.
Does anyone have any suggestions on how I can continue from here? Thanks in advance!
 A: Let $y=x-1$.  Then, we can write
$$\frac{x^k-1}{x-1}=\frac{(1+y)^k-1}{y} \tag1$$
Applying the binomial theorem to the right-hand side of $(1)$ reveals that 
$$\begin{align}
\left|\frac{x^k-1}{x-1}-k\right|&=\left|\sum_{\ell=2}^k \binom{k}{\ell}y^{\ell-1}\right|\\\\
&\le |y|\sum_{\ell=2}^k\binom{k}{\ell}|y|^{\ell-2}\tag 2
\end{align}$$
Now, we restrict $|y|$ such that $|y|\le 1$.  Then, we see that 
$$\sum_{\ell=2}^k\binom{k}{\ell}|y|^{\ell-2}\le 2^k$$
Therefore, given $\epsilon>0$ we have 
$$\left|\frac{x^k-1}{x-1}-k\right|<\epsilon$$
whenever $|x-1|<\delta=\min\left(1,\epsilon/2^k\right)$.
A: Hint: We can use $\frac{x^k -1}{x-1} = x^{k-1} + \ldots + x + 1$. Then together with the triangle inequality we find:
\begin{align}
  \Big|\frac{x^k -1}{x-1} - k\Big|
  &= |x^{k-1} + \ldots + x + 1 - k| \\
  &  \le \sum_{j=1}^{k-1} |x^j - 1| \\
  &= |x-1|\sum_{j=1}^{k-1}|1+x+\ldots+x^{j-1}| \\
  &<\delta\cdot \sum_{j=1}^{k-1}|1+x+\ldots+x^{j-1}|
\end{align}
And the last sum is obviously bounded if we request without loss of generality that e.g. $\delta\le 1\implies x\in[0,2]$
