Rational Functions and End Behavior

Worked example

1. 01

   Identify $n$ and $m$: the numerator has degree $n=1$, the denominator has degree $m=2$.

2. 02

   Since $n < m$, the denominator dominates — the fraction shrinks toward 0 in both directions.

3. 03

   Write the limit statements:

   $$\lim_{x \to -\infty} f(x) = 0 \qquad \lim_{x \to \infty} f(x) = 0$$

   The same idea works with any variable name — rational functions don't have to use $x$ and $f$. For example:

   $$w(k) = \frac{3k - 5}{2k}$$

   Here n = $m = 1$ (equal degrees), so neither dominates — the limit is the ratio of leading coefficients:

   $$\lim_{k \to \pm\infty} w(k) = \frac{3}{2}$$

Worked example

1. 01

   You don't need to fully expand to find the degree — just multiply the leading terms: (x)(3x) = $3x^{2}$, so the numerator is degree 2. But let's expand fully to also get the leading coefficient right:

   $$(x - 3)(3x + 2) = 3x^2 - 7x - 6 \;\text{ — degree 2}$$
2. 02

   Now compare: numerator degree 2, denominator degree 2 $(x^{2}-16)$ — equal degrees, neither dominates.

3. 03

   Take the ratio of leading coefficients:

   $$\lim_{x \to \infty} \frac{3x^2 - 7x - 6}{x^2 - 16} = \frac{3}{1} = 3$$

   One more — this time the numerator dominates:

   $$\lim_{x \to -\infty} \frac{4x^3 - 5}{x^2 + 10}$$ $$\text{degree } 3 > \text{degree } 2 \;\Rightarrow\; \text{no horizontal asymptote}$$

Worked example

1. 01

   Confirm this is the slant-asymptote case: numerator degree (2) is exactly one more than denominator degree (1).

2. 02

   Divide using the standard long-division layout:

   $$\begin{array}{r} x\phantom{{}+{}1} \\ x\,\overline{\smash{\big)}\,x^2+1\phantom{{}}} \\ \underline{-\;x^2\phantom{{}+{}1}} \\ 1 \end{array}$$
3. 03

   So $x^{2} + 1$ divided by $x$ gives a quotient of $x$, with a remainder of 1:

   $$f(x) = \frac{x^2 + 1}{x} = x + \frac{1}{x}$$
4. 04

   As $x$ grows large:

   $$\text{As } x \to \pm\infty, \;\; \frac{1}{x} \to 0$$
5. 05

   The slant asymptote is $y = x$ — matching the graph above.

Worked example

1. 01

   Confirm the case: numerator degree 2 is one more than denominator degree 1.

2. 02

   Divide using the standard long-division layout:

   $$\begin{array}{r} 2x+1\phantom{{}-{}1} \\ x+1\,\overline{\smash{\big)}\,2x^2+3x-1} \\ \underline{-\;(2x^2+2x)\phantom{{}-{}1}} \\ x-1 \\ \underline{-\;(x+1)} \\ -2 \end{array}$$
3. 03

   Rewrite as quotient plus remainder-over-divisor:

   $$g(x) = 2x + 1 - \frac{2}{x + 1}$$
4. 04

   The quotient is 2x + 1, so the slant asymptote is $y = 2x + 1$.

Worked example

1. 01

   The quotient is $2x^{2} - 7x + 18$, with a remainder of −31.

2. 02

   Since the remainder over $(x+2)$ shrinks to 0 as $x \to \pm \infty$, the function behaves like $y = 2x^{2} - 7x + 18$ far from the origin — a parabola, not a line, because the quotient itself is degree 2.

Worked example

1. 01

   Numerator degree is 3, denominator degree is 1 — a difference of 2, not 1.

2. 02

   No slant asymptote exists in this case (the end behavior instead resembles a parabola, not a line).

Worked example

1. 01

   Divide:

   $$x^2 - 4 = (x - 1)(x + 1) - 3$$
2. 02

   Rewrite:

   $$h(x) = x + 1 - \frac{3}{x - 1}$$
3. 03

   The slant asymptote is $y = x + 1$.