Problem 595
Easy Tangents, Normals, and Asymptotes

Find all asymptotes of y=x sqrt((x-1)/(x-2))

Find all asymptotes of $y=x\sqrt{\dfrac{x-1}{x-2}}$.

Explanation of Method

Rationalizing a square-root difference again produces the oblique intercept in closed form.

Step 1: Domain and vertical asymptote

The radicand is $\ge0$ for $x\le1$ or $x>2$. At $x=2$, the numerator $x-1\to1$ while the denominator $\to0^+$, so

\[ \lim_{x\to2^{+}} x\sqrt{\frac{x-1}{x-2}} = +\infty \;\Longrightarrow\; x=2\ \text{is a vertical asymptote.} \tag{1} \]

Step 2: Oblique asymptote — slope

\[ k = \lim_{x\to\infty}\sqrt{\frac{x-1}{x-2}} = \sqrt{\lim_{x\to\infty}\frac{1-1/x}{1-2/x}} = 1. \tag{2} \]

Step 3: Oblique asymptote — intercept

Rationalizing: \[\sqrt{\frac{x-1}{x-2}}-1 = \frac{1}{(x-2)\!\left(\sqrt{\dfrac{x-1}{x-2}}+1\right)}, \tag{3}\]

\[ b = \lim_{x\to\infty} x\!\left(\sqrt{\frac{x-1}{x-2}}-1\right) = \lim_{x\to\infty}\frac{x}{(x-2)\!\left(\sqrt{\dfrac{x-1}{x-2}}+1\right)} = \frac1{1\cdot2} = \frac12. \tag{4} \]

Step 4: Verification

\[x=1000\text{: } y-(x+0.5)\approx0.00025\to0;\qquad x\to2^{+}\text{: }y\to+\infty\ \checkmark\]


Result

Final answer:

\[\boxed{\,x = 2\ \text{(vertical)}; \qquad y = x+\frac12\ \text{(oblique)}\,}\]