$\lim_{x \to 1} \frac{x^k - 1}{x - 1}$ using $\delta, \epsilon$

Question

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!

Answer 1

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 $|x-1|<\delta=\min\left(1,\epsilon/2^k\right)$.

Answer 2

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. $\delta\le 1\implies x\in[0,2]$