Limits of rational functions often require more than direct substitution. If substitution gives a finite value,
the limit is immediate. If substitution gives an indeterminate form such as \(\frac00\), factoring or rationalizing
can reveal the nearby behavior.
1. Direct substitution
For a rational function:
\[
f(x)=\frac{P(x)}{Q(x)}
\]
if \(Q(a)\ne0\), then:
\[
\lim_{x\to a}\frac{P(x)}{Q(x)}
=
\frac{P(a)}{Q(a)}
\]
2. The indeterminate form \(\frac00\)
If direct substitution gives:
\[
\frac{0}{0}
\]
the expression is indeterminate. This does not mean the limit is zero or undefined automatically.
It means more algebra is needed.
3. Factoring and canceling common factors
A common situation is:
\[
\frac{(x-a)P(x)}{(x-a)Q(x)}
\]
For \(x\ne a\), the common factor can be canceled:
\[
\frac{(x-a)P(x)}{(x-a)Q(x)}
=
\frac{P(x)}{Q(x)}
\]
Then the limit can be evaluated using the simplified expression:
\[
\lim_{x\to a}\frac{(x-a)P(x)}{(x-a)Q(x)}
=
\lim_{x\to a}\frac{P(x)}{Q(x)}
\]
4. Worked example: factoring
Evaluate:
\[
\lim_{x\to2}\frac{x^2-4}{x-2}
\]
Direct substitution gives:
\[
\frac{2^2-4}{2-2}
=
\frac00
\]
Factor the numerator:
\[
x^2-4=(x-2)(x+2)
\]
Cancel the common factor:
\[
\frac{x^2-4}{x-2}
=
\frac{(x-2)(x+2)}{x-2}
=
x+2,
\qquad x\ne2
\]
Now substitute:
\[
\lim_{x\to2}\frac{x^2-4}{x-2}
=
\lim_{x\to2}(x+2)
=
4
\]
5. Removable discontinuities
In the previous example, the original function is undefined at \(x=2\), but the limit exists.
This creates a removable discontinuity, also called a hole.
\[
f(2)\text{ is undefined, but }\lim_{x\to2}f(x)=4
\]
6. Rationalizing with a conjugate
Radical expressions can also produce \(\frac00\). For example:
\[
\lim_{x\to2}\frac{\sqrt{x+2}-2}{x-2}
\]
Direct substitution gives \(\frac00\). Multiply by the conjugate:
\[
\frac{\sqrt{x+2}-2}{x-2}
\cdot
\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}
\]
Use the difference of squares:
\[
(\sqrt{x+2}-2)(\sqrt{x+2}+2)
=
(x+2)-4
=
x-2
\]
Then:
\[
\frac{\sqrt{x+2}-2}{x-2}
=
\frac{x-2}{(x-2)(\sqrt{x+2}+2)}
=
\frac{1}{\sqrt{x+2}+2}
\]
Finally:
\[
\lim_{x\to2}\frac{\sqrt{x+2}-2}{x-2}
=
\frac{1}{\sqrt4+2}
=
\frac14
\]
7. Common algebraic patterns
| Pattern |
Method |
Example |
| Difference of squares |
Factor and cancel |
\(\frac{x^2-a^2}{x-a}\) |
| Quadratic trinomial |
Factor and cancel |
\(\frac{x^2+bx+c}{x-a}\) |
| Difference of cubes |
Factor and cancel |
\(\frac{x^3-a^3}{x-a}\) |
| Radical difference |
Multiply by conjugate |
\(\frac{\sqrt{x+h}-c}{x-a}\) |
| Nonzero denominator |
Direct substitution |
\(\frac{x^2+1}{x+3}\) |
8. Infinite behavior
If the denominator approaches \(0\) but the numerator does not, the limit may be infinite or may not exist.
For example:
\[
\lim_{x\to2}\frac{1}{x-2}
\]
From the left, the expression approaches \(-\infty\). From the right, it approaches \(+\infty\).
Therefore, the two-sided limit does not exist.
9. Common mistakes
- Do not cancel terms that are added or subtracted; cancel only common factors.
- Do not say \(\frac00=0\). The form \(\frac00\) is indeterminate.
- After canceling, remember that the original function may still be undefined at the canceled point.
- Use a conjugate when a square-root difference causes \(\frac00\).
- Always check left and right behavior if the denominator still approaches zero.
- A finite limit can exist even if \(f(a)\) is undefined.
