How to get asymptotes of hyperbola using derivatives?

Question

I'm trying to get the asymptotes of a hyperbola of the form $f(x)=\sqrt{1+px^2}$. So I took the derivative giving me:

$${d\over dx}\left(\sqrt{1+px^2} \right)={px\over \sqrt{1+px^2}} $$

Now, giving that the asymptote lines $T(a)$ at some point $(a,f(a))$ of the graph is:

$$T(a)=f'(a)(x-a)+f(a) $$

I decided to take the limit

$$\lim_{a\to\infty}T(a)=\lim_{a\to\infty}\left({pa\over \sqrt{1+pa^2}} \right)(x-a)+ \sqrt{1+pa^2}$$

I tried L'Hopital's rule and didn't work.

Answer 1

If a do a substitution $a=1/u$, such that taking the limit when $u\to 0$ then:

$$\lim_{a\to\infty}\left({pa\over \sqrt{1+pa^2}} \right)(x-a)+ \sqrt{1+pa^2}$$

$$=\lim_{u\to 0}\left({p(1/u)\over \sqrt{1+p(1/u)^2}} \right)(x-(1/u))+ \sqrt{1+p(1/u)^2} $$

$$=\lim_{u\to 0}\left({p\over \sqrt{u^2+p}} \right)\left(ux-1\over u \right)+{{\sqrt{u^2+p}}\over u}$$

$$=\lim_{u\to0}{{pux-p+u^2+p}\over u\sqrt{u^2+p}}=\lim_{u\to 0}{{px+u}\over\sqrt{u^2+p}} $$ $$={px\over \sqrt{p}} $$

And so$\dots$

$$\lim_{a\to\infty}T(a)=x\sqrt{p} $$

Answer 2

$$f(x)=\sqrt{1+px^2}=\sqrt{px^2\left(1+\frac{1}{px^2}\right)}=\sqrt{px^2}\sqrt{1+\frac{1}{px^2}}$$ Using the fact that $(1+t)^{\alpha}\sim1+\alpha t$ when $t\to 0$, and assuming $x>0$, you have: $$ f(x)\sim x\sqrt{p}\left(1+\frac{1}{2px^2}\right)=x\sqrt{p}+\frac{1}{2\sqrt{p}x}$$