If a function tends to a as x tends to infinity and to -a as x tends to minus infinity, how many horizontal asymptotes does it have?

Step-by-step solution

1. 1

   Recall the definition — each end of the graph is judged separately. The line $y=L$ is a horizontal asymptote of $f$ if $f(x)\to L$ as $x\to+\infty$ or as $x\to-\infty$. The two directions are independent conditions, which is why a single function is allowed more than one horizontal asymptote (unlike a vertical asymptote count, which is unlimited for a different reason).

2. 2

   Apply it to each limit. The limit $a$ at $+\infty$ makes $y=a$ an asymptote, and the limit $-a$ at $-\infty$ makes $y=-a$ an asymptote. Both are legitimate provided $a$ is a finite number — an infinite limit is no asymptote at all, and would instead suggest unbounded growth or a slant asymptote.

3. 3

   Count them, treating $a=0$ separately. If $a\neq0$ then $a$ and $-a$ are two different numbers, so there are two horizontal asymptotes:

   $y=a\qquad\text{and}\qquad y=-a$

   If $a=0$ then $a=-a=0$, the two lines coincide, and there is only one asymptote, $y=0$.

4. 4

   Work a concrete example. Take

   $f(x)=\frac{x}{\sqrt{x^{2}+1}}$

   For large $|x|$, $\sqrt{x^{2}+1}\approx|x|$, so $f(x)\approx x/|x|$, which is $+1$ for positive $x$ and $-1$ for negative $x$. Numerically $f(10^{8})=1$ and $f(-10^{8})=-1$ to within $10^{-12}$ ✓, so this function has the two asymptotes $y=1$ and $y=-1$ — the case $a=1$.

5. 5

   Note the maximum possible count. A function can have at most two horizontal asymptotes, one per direction, because each of $x\to+\infty$ and $x\to-\infty$ admits only a single limit. Standard examples include $\arctan x$ (with $y=\pm\pi/2$) and $\tanh x$ (with $y=\pm1$) — both exactly of the $a$, $-a$ shape.