Proving that $y=x$ is an asymptote.

Question

I'm currently investigating the curve implicitly defined by $x^2-y^2= \ln (xy)$. Now I can see that by considering the region $xy>1$ and $xy<1$, we can determine where the curve is above or below the line $y=x$.

However, I do not feel that this is not enough to prove that the line $y=x$ is an asymptote in itself, just whether it approaches it from above or below.

For an implicitly defined curve like this, what kind of approach would I take to show that $y=x$ is an asymptote?

Answer 1

You could separate the variables:

$$x^2 - \ln x = y^2 + \ln y.$$

It's easy then to show that $y\to\infty$ as $x\to\infty$.

Take the ratio of the two sides: $$\frac{y^2 + \ln y}{x^2 - \ln x} = 1.$$

For large positive $x$, you have large positive $y$ and $y^2/x^2 \approx 1$, so $y/x \approx 1$.

You can even write \begin{align} \frac{y^2}{x^2} & = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right) \frac{y^2 + \ln y}{x^2 - \ln x} \\ & = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right) \end{align}

Each of the quantities in parentheses goes to $1$ as $x\to\infty$.

Answer 2

You can establish that $$\frac{dy}{dx}=\frac{2x-\frac 1x}{2y+\frac 1y}\rightarrow 1$$ as $x,y\rightarrow +\infty$

Hence for large $x,y>0$, the graph is approximated by the line $y=x$