Given the curve y=xe^(1/x). Find the equation

Given the curve $y=xe^{1/x}$. Find the equation of the tangent at $x=\alpha$ and its limiting position as $\alpha\to\infty$.

Step 1: Derivative

By the product rule, \[y' = e^{1/x} + x\cdot e^{1/x}\!\left(-\frac1{x^{2}}\right) = e^{1/x}\!\left(1-\frac1x\right). \tag{1}\]

Step 2: Tangent at $x=\alpha$

The point of tangency is $P=\left(\alpha,\,\alpha e^{1/\alpha}\right)$ with slope $k=e^{1/\alpha}\left(1-\tfrac1\alpha\right)$; expanding the point–slope form, the constant term $-\alpha e^{1/\alpha}(1-\tfrac1\alpha)+\alpha e^{1/\alpha}$ collapses to $e^{1/\alpha}$, giving

\[y = e^{1/\alpha}\!\left(1-\frac1\alpha\right)x + e^{1/\alpha}. \tag{2}\]

Step 3: Limiting position as $\alpha\to\infty$

Since $\displaystyle\lim_{\alpha\to\infty}e^{1/\alpha}\!\left(1-\frac1\alpha\right)=1$ and $\displaystyle\lim_{\alpha\to\infty}e^{1/\alpha}=1$, (2) tends to

\[ y = x+1. \tag{3} \]

Step 4: Geometric interpretation

Indeed $y=x+1$ is the oblique asymptote of $y=xe^{1/x}$, since, with $u=1/x\to0^{+}$ and $e^{u}-1-u=\tfrac{u^2}2+\cdots$, \[\lim_{x\to\infty}\bigl(xe^{1/x}-x-1\bigr) = \lim_{x\to\infty}x\!\left(e^{1/x}-1-\frac1x\right) = \lim_{u\to0^{+}}\frac{e^{u}-1-u}{u} = 0. \tag{4}\]

The limiting tangent is exactly the oblique asymptote of the curve.

Step 5: Verification

\[\text{Tangent at } x=\alpha\text{: } e^{1/\alpha}\!\left(1-\tfrac1\alpha\right)\alpha+e^{1/\alpha}=e^{1/\alpha}(\alpha-1+1)=\alpha e^{1/\alpha}\ \checkmark\]

\[\text{Both coefficients} \to 1 \text{ as } \alpha\to\infty,\text{ giving } y=x+1\ \checkmark\]