Proving the surprising limit: $\lim\limits_{n \to 0} \frac{x^{n}-y^{n}}{n}$ $=$ log$\frac{x}{y}$

A few months ago, while at school, my classmate asked me this curious question: What does $\frac{x^{n}-y^{n}}{n}$ tend to as $n$ tends to $0$? I thought for a few minutes, became impatient, and asked "What?" His reply, log$\frac{x}{y}$, was surprising, but his purported 'proof' was more surprising:

Consider $\lim\limits_{n \to 0}\,\int_y^x t^{n-1}\, dt$. "Pushing the limit into the definite integral", we have $$\int_y^x \lim\limits_{n \to 0}\,t^{n-1}\, dt \implies \int_y^x \frac{1}{t}\, dt \implies \mathsf{log} \frac{x}{y}$$

Leaving the fact that he had the inspiration to pull this integral out of thin air aside, is the limit allowed to pass into the definite integral? We hadn't learned Real Analysis (we were just taking a basic high school, hand-wavy single-variable calculus course), and I remember feeling very uneasy about the sorcery. I still am, hence, this question.

I've since thought about approaching it using $\mathsf{L'Hospital}$, but I still feel uneasy, since it involves differentiating with respect to different variables, which is a little bit confusing. I'd also appreciate your help in this regard.

If you have a better proof, I'll truly appreciate it.

• Actually, you don't need to differentiate with respect to different variables since $x$ and $y$ are constants here (unless you mean not differentiating with respect to $x$ as is most commonly done). Alternatively, you may recognize the limit as the derivative of the function $f(t) = x^t-y^t$ at $t=0$. Commented Jan 24, 2016 at 7:25
• Right; now it remains to prove that the integral 'sorcery' is valid. Commented Jan 24, 2016 at 7:29

By using L'Hospital rule we have \begin{align*} \lim_{n\to 0}\frac{x^n-y^n}{n}&=\lim_{n\to 0}\frac{x^n\log x-y^n\log y}{1}\\ &=\log x-\log y\\&=\log\frac{x}{y} \end{align*}

• You never need L'Hospital's Rule when the denominator is h itself since in that case the limit is just the definition of the derivative at that point Commented May 8, 2023 at 7:36

In general, passing the limit under the integral is not valid. In this circumstance, however, it is valid, as the functions converge uniformly on a finite interval. Uniform convergence means the following: if we let $M_n = \max\limits_{y\le t\le x}{|t^{n-1}-t^{-1}|}$ be the maximum difference of $t^{n-1}$ from $t^{-1}$, then $\lim\limits_{n\rightarrow 0}{M_n} = 0$. You can check that uniform convergence does hold here, so by a standard (and fairly easy to prove) theorem of analysis, passing the limit under the integral is justified.

Hint: \begin{align} \lim_{n\to0}\frac{x^n-1}{n} &=\log(x)\lim_{n\to0}\frac{e^{n\log(x)}-1}{n\log(x)}\\ &=\log(x)\lim_{u\to0}\frac{e^u-1}u\\[6pt] &=\log(x) \end{align} Subtract

Swapping Integral and Limit

Let $$\lambda_{\rm{min}}=\min\left\{1,y^{-5/4}\right\}$$ and $$\lambda_{\rm{max}}=\max\left\{1,x^{-5/4}\right\}$$, then for $$n\in\left[-\frac14,\frac14\right]$$, the Mean Value Theorem says that for some $$\tau$$ between $$1$$ and $$t$$, we have $$\frac{t^n-1}{t-1}=n\tau^{n-1}=n\,\left[\lambda_{\rm{min}},\lambda_{\rm{max}}\right]_\#$$ where $$[a,b]_\#$$ represents a number in $$[a,b]$$. \begin{align} \left|\int_x^y\left(t^{n-1}-t^{-1}\right)\mathrm{d}t\right| &=\left|\int_x^y\left(t^n-1\right)t^{-1}\,\mathrm{d}t\right|\\ &=\left|n\int_x^y\left[\lambda_{\rm{min}},\lambda_{\rm{max}}\right]_\#\,(t-1)\,t^{-1}\,\mathrm{d}t\right|\\ &\le|n|\,\underbrace{\lambda_{\rm{max}}\int_x^y\,|t-1|\,t^{-1}\,\mathrm{d}t}_\text{a constant} \end{align} Thus, in this case, it is valid to swap the integral and the limit $$\lim_{n\to0}\int_x^yt^{n-1}\,\mathrm{d}t=\int_x^yt^{-1}\,\mathrm{d}t$$

Let $t = x/y$ and notice that \begin{align*} \lim_{n \to 0} \frac{x^n - y^n}{n} &= \lim_{n \to 0} y^n \frac{t^n - 1}{n} \\ &= \left(\lim_{n \to 0} y^n\right) \left(\lim_{n \to 0} \frac{t^n - 1}{n}\right) \\ &= 1 \cdot \frac{d}{dn}\Big|_{n = 0} t^n \\ &= \ln t = \ln \frac x y \end{align*}

as desired.

\begin{align} \lim_{n\to0}\frac{x^n-y^n}{n}&=\lim_{n\to0}\frac{e^{n\cdot \ln x}-e^{0\cdot \ln x}-(e^{n\cdot \ln y}-e^{0\cdot \ln y})}{n}\\ &=\left(e^{n\cdot \ln x}\right)'|_{n=0}-\left(e^{n\cdot \ln y}\right)'|_{n=0}\\ &=\ln x - \ln y \end{align}