Basic limit laws let us evaluate many limits by breaking a complicated expression into simpler pieces.
When the function is continuous at the approach value, direct substitution gives the limit immediately.
1. Direct substitution
If \(f\) is continuous at \(x=a\), then:
\[
\lim_{x\to a}f(x)=f(a)
\]
This means we can replace \(x\) by \(a\), as long as the expression remains defined.
2. Example: direct substitution
Evaluate:
\[
\lim_{x\to4}(2x+3)
\]
The function \(2x+3\) is linear, so it is continuous everywhere. Substitute \(x=4\):
\[
2(4)+3=8+3=11
\]
Therefore:
\[
\lim_{x\to4}(2x+3)=11
\]
3. Constant law
A constant approaches itself:
\[
\lim_{x\to a}c=c
\]
For example:
\[
\lim_{x\to5}7=7
\]
4. Identity law
The identity function \(f(x)=x\) approaches the input value:
\[
\lim_{x\to a}x=a
\]
5. Sum and difference laws
If the limits of \(f(x)\) and \(g(x)\) both exist, then:
\[
\lim_{x\to a}\bigl(f(x)+g(x)\bigr)
=
\lim_{x\to a}f(x)+\lim_{x\to a}g(x)
\]
\[
\lim_{x\to a}\bigl(f(x)-g(x)\bigr)
=
\lim_{x\to a}f(x)-\lim_{x\to a}g(x)
\]
6. Constant multiple law
A constant factor can be pulled outside the limit:
\[
\lim_{x\to a}c f(x)
=
c\lim_{x\to a}f(x)
\]
For example:
\[
\lim_{x\to2}3x
=
3\lim_{x\to2}x
=
3(2)=6
\]
7. Product law
If both separate limits exist, then:
\[
\lim_{x\to a}\bigl(f(x)g(x)\bigr)
=
\left(\lim_{x\to a}f(x)\right)
\left(\lim_{x\to a}g(x)\right)
\]
For example:
\[
\lim_{x\to2}(x+2)(3x-1)
=
(2+2)(3\cdot2-1)
=
4\cdot5=20
\]
8. Quotient law
If both limits exist and the denominator limit is not zero, then:
\[
\lim_{x\to a}\frac{f(x)}{g(x)}
=
\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)},
\qquad
\lim_{x\to a}g(x)\ne0
\]
The condition \(\lim_{x\to a}g(x)\ne0\) is essential.
If the denominator limit is \(0\), direct substitution cannot use the quotient law.
9. Power and root laws
Powers can usually be handled by substitution:
\[
\lim_{x\to a}\bigl(f(x)\bigr)^n
=
\left(\lim_{x\to a}f(x)\right)^n
\]
Roots are also continuous on their domains. For example:
\[
\lim_{x\to4}\sqrt{x+5}
=
\sqrt{4+5}
=
3
\]
10. Continuous functions
Many common functions are continuous on their domains, so direct substitution works:
| Function type |
Continuity condition |
Substitution rule |
| Polynomial |
Continuous for all real \(x\) |
\(\lim_{x\to a}P(x)=P(a)\) |
| Rational |
Continuous where denominator is not \(0\) |
\(\lim_{x\to a}\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}\), if \(Q(a)\ne0\) |
| Root |
Expression under even root must be nonnegative |
\(\lim_{x\to a}\sqrt{f(x)}=\sqrt{f(a)}\), if defined |
| Logarithm |
Input must be positive |
\(\lim_{x\to a}\ln(f(x))=\ln(f(a))\), if \(f(a)\gt0\) |
| Trigonometric |
Continuous where defined |
Substitute directly when the expression is defined |
11. When direct substitution fails
Direct substitution fails when the expression becomes undefined. For example:
\[
\lim_{x\to1}\frac{x^2-1}{x-1}
\]
Direct substitution gives:
\[
\frac{1^2-1}{1-1}
=
\frac{0}{0}
\]
The form \(\frac00\) is indeterminate. Basic direct substitution is not enough.
You need another method, such as factoring or cancellation.
12. Law-by-law example
Evaluate:
\[
\lim_{x\to3}\frac{x^2+1}{x+2}
\]
The numerator and denominator are continuous:
\[
\lim_{x\to3}(x^2+1)=3^2+1=10
\]
\[
\lim_{x\to3}(x+2)=3+2=5
\]
Since the denominator limit is not zero, use the quotient law:
\[
\lim_{x\to3}\frac{x^2+1}{x+2}
=
\frac{10}{5}=2
\]
13. Common mistakes
- Do not use the quotient law if the denominator limit is \(0\).
- Do not cancel terms before checking that the original expression is defined.
- Do not assume every function is continuous everywhere.
- For logarithms, make sure the input is positive.
- For square roots, make sure the input is in the allowed domain.
- When direct substitution gives \(\frac00\), use a different limit tool.
