Asymptote for $y=±\sqrt{x^2-1}$

Question

Currently, i am working with hyperbolic functions, and I proved that $\cosh^2(x)-\sinh^2(x)=1$

Now from this we can make the parametric equation $(x,y)=\cosh(t) , \sinh(t)$ , $t\inℝ$

Then we can say, $x^2-y^2=1$ which leads to $y=\pm\sqrt{x^2-1}$

Now we can say that if $x$ is big and only positive values are of our interest $y=\sqrt{x^2-1}\approx\sqrt{x^2}=x$

Basically, the oblique asymptote is $y=x$

However, the definition for oblique asymptotes states that if

$\lim_{x\to∞}(f(x)-mx+n)=0$ or $\lim_{x\to-∞}(f(x)-mx+n)=0$

then the linear function $mx+n$ is the oblique asymptote.

In this case, is it then correct to argue that $\lim_{x\to∞}(\sqrt{x^2-1})≈x$ makes $x$ the oblique asymptote?

Answer 1

No, it is not correct. Writing $$ \lim_{x\to \infty} f(x) = x $$ makes no sense: a limit is "just a number", here you say it's a variable. This is not mathematically correct.

However, you can make this correct.

• A first method would be to write, for $x>0$, $$ \sqrt{x^2-1} - x = x\sqrt{1-1/x^2} -x $$ and use Taylor expansion of $\sqrt{1-u}$ at $0$ to show this converges to $0$. This is correct, but requires to know about Taylor expansions.

• Another, more elementary approach, is to use multiplication by the conjugate. You have $$\lvert \sqrt{x^2-1}-x \rvert = \left\lvert\frac{x^2-1-x^2}{\sqrt{x^2-1}+x} \right\rvert = \frac{1}{\sqrt{x^2-1}+x} \xrightarrow[x\to\infty]{} 0$$ since the denominator goes to $\infty$. To get the first equality, we used $$\sqrt{a}-\sqrt{b} = \frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}= \frac{a-b}{\sqrt{a}+\sqrt{b}}\,,$$ for $a,b\geq 0$.

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Note: As an example of why we need the limit of the difference to go to $0$ (i.e., $\lim_{x\to \infty} \lvert f(x) - x\rvert = 0$) instead of a more "fuzzy" $f(x) \approx x$", consider $f(x) = x+\sqrt{x}$. Surely, we also have $f(x)\approx x$ for some reasonable sense of $\approx$; yet then we wouldn't have an asymptote.