Prove that the curve x=(1+ dfrac1u)^u, y=(1+

Prove that the curve $x=\left(1+\dfrac1u\right)^{u},\ y=\left(1+\dfrac1u\right)^{u+1}$ intersects the line $y=x$ at right angles.

Step 1: The relation between $x$ and $y$

\[ y = \left(1+\frac1u\right)^{u+1} = x\!\left(1+\frac1u\right). \tag{1} \]

Step 2: Where the curve meets $y=x$

Setting $y=x$ in (1) forces $1+1/u=1$, i.e.\ $u\to\infty$, where $x,y\to e$. The intersection point is $(e,e)$.

Step 3: Slope as $u\to\infty$

Taking logarithmic derivatives of $x=v^{u}$, $y=v^{u+1}$ with $v=1+1/u$ (so $\dot v=-1/u^2$): \[\frac{\dot x}{x} = \ln v - \frac{1}{u+1}, \qquad \frac{\dot y}{y} = \ln v - \frac{1}{u}. \tag{2}\]

With $\varepsilon=1/u\to0^{+}$, Taylor expansion gives $\ln(1+\varepsilon)-\varepsilon=-\varepsilon^2/2+O(\varepsilon^3)$ and $\ln(1+\varepsilon)-\dfrac{\varepsilon}{1+\varepsilon}=\varepsilon^2/2+O(\varepsilon^3)$, so

\[ \frac{dy}{dx} = \frac yx\cdot\frac{\dot y/y}{\dot x/x} \;\longrightarrow\; 1\cdot\frac{-1/2}{1/2} = -1 \qquad (u\to\infty). \tag{3} \]

Step 4: Perpendicularity

The line $y=x$ has slope $1$; the curve's limiting slope is $-1$; the product is $-1$.

Step 5: Verification

\[u=1000\text{: numerically computed slope} \approx-1.00200,\text{ approaching }-1\text{ as } u\to\infty\ \checkmark\]