Connecting Limits at Infinity and Horizontal Asymptotes

Evaluate $\dfrac{3x^2}{x^2 + 1}$ at $x = 10$ and you get about $2.970$; at $x = 100$, about $2.9997$. As $x$ grows, the $+1$ in the denominator matters less and less, and the values settle toward 3. Topic 1.14 asked what happens as $x$ approaches a number; this topic asks what happens as $x$ itself grows without bound, which is the question of a function's end behavior: what its graph does at the far right and far left.

Limits at infinity

Writing

means that $f(x)$ stays as close to $L$ as you want once $x$ is large enough. Choose a tolerance like $0.001$, and from some point on, every value of $f(x)$ is within $0.001$ of $L$. The limit as $x \to -\infty$ means the same for inputs that are very negative. A function can also grow without bound at the ends, written $\lim\limits_{x \to \infty} f(x) = \infty$ (as $x^2$ does), or have no limit at all because it keeps oscillating (as $\sin x$ does, swinging between $-1$ and $1$ forever).

A few limits are the building blocks for everything else:

The first row is the workhorse: a fixed number divided by something growing without bound shrinks to 0. The last row is a function with two different horizontal asymptotes, $y = \frac{\pi}{2}$ on the right and $y = -\frac{\pi}{2}$ on the left.

A horizontal asymptote only describes what happens far out. Unlike a vertical asymptote, it can be crossed, any number of times. The function $g(x) = \dfrac{2x^2 + 3x}{x^2 + 1}$ has $\lim\limits_{x \to \pm\infty} g(x) = 2$, yet $g(x) = 2$ when $3x = 2$, so the graph crosses $y = 2$ at $x = \frac{2}{3}$, rises above it, and then falls back toward it.

Rational functions: the leading terms decide

For large $x$, the highest power in a polynomial dwarfs everything else. At $x = 1000$, the term $5x$ is about $0.17\%$ of $3x^2$. So the end behavior of a rational function comes down to the leading term of the numerator over the leading term of the denominator. To show it with limits, divide every term by the highest power of $x$ in the denominator, then use $\frac{c}{x^p} \to 0$:

A rational function always has the same limit at both ends (when the limit is finite), so it has at most one horizontal asymptote.

Radicals, and why $x \to -\infty$ needs care

With a square root, the dominant term is still the one with the highest power, but $\sqrt{x^2} = |x|$, not $x$. For negative $x$, $|x| = -x$, and that sign flip is where two different asymptotes come from. Take $\dfrac{\sqrt{4x^2 + 1}}{x - 3}$. Inside the root, factor out $x^2$: $\sqrt{4x^2 + 1} = |x|\sqrt{4 + \frac{1}{x^2}}$. Then

The graph has two horizontal asymptotes, $y = 2$ and $y = -2$. The sign check is reasonable: for very negative $x$, the numerator is positive and the denominator is negative.

Comparing how fast functions grow

When $f$ and $g$ both grow without bound, the limit of their ratio says which one grows faster:

"Faster" is about the long run, not about which function is bigger at a particular $x$. At $x = 10$, $x^{10} = 10^{10}$ is far bigger than $e^{10} \approx 22{,}026$. But $e^x$ overtakes $x^{10}$ near $x \approx 35.8$ and stays ahead, and by $x = 100$ it is about $2.7 \times 10^{43}$ against $x^{10} = 10^{20}$. Each step from $x$ to $x + 1$ multiplies $e^x$ by $e \approx 2.718$, while it multiplies $x^{10}$ by $\left(\frac{x+1}{x}\right)^{10}$, a factor that sinks toward 1 as $x$ grows. A constant multiplier always beats one that shrinks toward 1. The proof that the ratio goes to 0 uses L'Hospital's Rule (topic 4.7); until then, use the ordering as a known fact.

"Same rate" can hide a large difference. $x^2 + 100x$ is always bigger than $x^2$ for $x > 0$, by $100x$, which itself grows without bound. Still, $\dfrac{x^2 + 100x}{x^2} = 1 + \dfrac{100}{x} \to 1$, so the two grow at the same rate: the extra $100x$ is a vanishing fraction of the total.

Taking the limit of a ratio works the same way for any two quantities that both go to infinity, or both go to 0, including two rates of change. If two tanks drain at rates that both approach 0, the limit of the ratio of the rates tells you which tank's flow becomes negligible next to the other's. Example 4 does this.

Worked examples

Where students lose points

On the exam

A horizontal asymptote is justified by a limit statement, such as $\lim\limits_{x \to \infty} C(t) = 20$, for each direction you claim. When a question gives a function in context, it can ask for the limit as $t \to \infty$ with an interpretation; the interpretation names the quantity, its units, and the long-run value, as in Example 4(a).