I'm trying to determine the limit for $x\to \infty$ of the following expression: $$ x-\frac{x-1}{x}2e^{-1/x}K_1(2/x)=x \left[1- \frac{x-1}{x^2}2e^{-1/x}K_1(2/x) \right], $$ where $K_1(\cdot)$ denote the modified Bessel function (e.g., https://dlmf.nist.gov/10.25). My attempt:
Using the limiting form in 10.30.2 provided at https://dlmf.nist.gov/10.30, I would obtain \begin{equation} \begin{split} 2e^{-1/x}K_1(2/x) &\sim 2e^{-1/x} \frac{1}{2}\Gamma(1)\left(\frac{2/x}{2} \right)^{-1}\\ &=e^{-1/x} x\\ &\sim \left(1-\frac{1}{x} \right) x \end{split} \end{equation} as $x \to \infty$. This would leave me with \begin{equation} \begin{split} x \left[1- \frac{x-1}{x^2}2e^{-1/x}K_1(2/x) \right]&\sim x \left[1- \frac{x-1}{x^2}\left(1-\frac{1}{x} \right) x\right]\\ &=x\left[1- \frac{1}{x}\left(1-\frac{1}{x} \right)^2 x\right]\\ &= x \left[ \frac{2}{x}-\frac{1}{x^2}\right]\\ &\to 2. \end{split} \end{equation}
QUESTION: Is the above result correct? My intuition and some attempts to plot the function would have suggested me that the limit is $0$, so I'm wondering whether there's something I'm missing in the above reasoning. Any help is appreciated.
Let $x=\frac 2y$ to make the expression $$\frac{2}{y}+e^{-y/2} (y-2) K_1(y)$$ Expanding around $y=0$ gives $$2-y \left(\log (y)+\gamma +\frac{1}{4}-\log (2)\right)+O\left(y^2\right)$$