Computing limits
Tables and graphs help us estimate limits. But we can find most limits exactly with algebra. This page shows how.
The limit laws
The limit laws tell us how limits work with addition, multiplication, and other operations.
Suppose the limits of $f(x)$ and $g(x)$ as $x\to a$ both exist. Then:
1. Sum law: The limit of a sum is the sum of the limits.
$\displaystyle\lim_{x\to a}[f(x)+g(x)]=\displaystyle\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$
2. Difference law: The limit of a difference is the difference of the limits.
$\displaystyle\lim_{x\to a}[f(x)-g(x)]=\displaystyle\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$
3. Constant multiple law: If $c$ is a constant, then $c$ can be moved outside the limit.
$\displaystyle\lim_{x\to a}[c\,f(x)]=c\lim_{x\to a}f(x)$
4. Product law: The limit of a product is the product of the limits.
$\displaystyle\lim_{x\to a}[f(x)\,g(x)]=\displaystyle\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)$
5. Quotient law: The limit of a quotient is the quotient of the limits, as long as the limit of the denominator is not $0$.
$\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$
This law needs $\displaystyle\lim_{x\to a}g(x)\neq 0$.
6. Power law: For any positive integer $n$,
$\displaystyle\lim_{x\to a}[f(x)]^n=\left[\lim_{x\to a}f(x)\right]^n$
7. Root law: For any positive integer $n$,
$\displaystyle\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\displaystyle\lim_{x\to a}f(x)}$
If $n$ is even, we also need $\displaystyle\lim_{x\to a}f(x)>0$.
We also use two basic limits:
$\displaystyle\lim_{x\to a}c=c$ (the limit of a constant is that constant)
$\displaystyle\lim_{x\to a}x=a$ (as $x$ approaches $a$, $x$ approaches $a$)
Example 1: Use the limit laws to find $\displaystyle\lim_{x\to 2}(3x^2+5x-4)$.
Solution:
Use the sum, difference, and constant multiple laws to split the limit:
$\begin{align*}&\lim_{x\to 2}(3x^2+5x-4)\\&=3\lim_{x\to 2}x^2+5\lim_{x\to 2}x-\lim_{x\to 2}4\end{align*}$
Now use the power law and the two basic limits:
$\begin{align*}&=3(2)^2+5(2)-4\end{align*}$
Simplify:
$\begin{align*}&\lim_{x\to 2}(3x^2+5x-4)=18\end{align*}$
Direct substitution
Look at the answer in Example 1. It is the same as the value of the polynomial at $x=2$:
$3(2)^2+5(2)-4=18$.
This is always true for polynomials. It gives us a shortcut, called direct substitution.
Direct substitution: If $p(x)$ is a polynomial, then
$\displaystyle\lim_{x\to a}p(x)=p(a)$.
It also works for a rational function (a fraction of two polynomials), as long as the denominator is not $0$ at $a$:
$\displaystyle\lim_{x\to a}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)}$, if $q(a)\neq 0$.
It also works for roots of these functions, as long as the root is a real number at $x=a$.
Example 2: Find $\displaystyle\lim_{x\to -1}\dfrac{x^2+4}{x+3}$.
Solution:
At $x=-1$, the denominator is $-1+3=2$. It is not $0$, so we can use direct substitution:
$\begin{align*}&\lim_{x\to -1}\dfrac{x^2+4}{x+3}\\&=\dfrac{(-1)^2+4}{-1+3}\\&=\dfrac{5}{2}\end{align*}$
Example 3: Find $\displaystyle\lim_{x\to 4}\sqrt{2x+1}$.
Solution:
At $x=4$, the expression under the square root is $2(4)+1=9$. It is positive, so we can use direct substitution:
$\begin{align*}&\lim_{x\to 4}\sqrt{2x+1}\\&=\sqrt{2(4)+1}\\&=\sqrt{9}\\&=3\end{align*}$
When direct substitution gives 0/0
Sometimes direct substitution gives $\dfrac{0}{0}$. For example, try
$\displaystyle\lim_{x\to 3}\dfrac{x^2-9}{x-3}$.
Substituting $x=3$ gives $\dfrac{9-9}{3-3}=\dfrac{0}{0}$.
The expression $\dfrac{0}{0}$ is called an indeterminate form. It does not tell us the limit. The limit could be any number, or it may not exist. So we first change the form of the function, using algebra. Then we try direct substitution again.
Why can we change the function? In a limit, $x$ approaches $a$ but never equals $a$. So a factor like $x-3$ is never $0$ while $x$ approaches $3$. This means we can cancel it.
Method 1: Factor and cancel
Example 4: Find $\displaystyle\lim_{x\to 3}\dfrac{x^2-9}{x-3}$.
Solution:
Direct substitution gives $0/0$. Factor the numerator as a difference of squares:
$\begin{align*}&\lim_{x\to 3}\dfrac{x^2-9}{x-3}\\&=\lim_{x\to 3}\dfrac{(x-3)(x+3)}{x-3}\end{align*}$
Cancel the common factor $x-3$:
$\begin{align*}&=\lim_{x\to 3}(x+3)\end{align*}$
Now substitute $x=3$:
$\begin{align*}&\lim_{x\to 3}\dfrac{x^2-9}{x-3}=3+3=6\end{align*}$
Example 5: Find $\displaystyle\lim_{x\to 2}\dfrac{x^2+x-6}{x^2-4}$.
Solution:
Direct substitution gives $0/0$. Factor the numerator and the denominator:
$\begin{align*}&\lim_{x\to 2}\dfrac{x^2+x-6}{x^2-4}\\&=\lim_{x\to 2}\dfrac{(x+3)(x-2)}{(x+2)(x-2)}\end{align*}$
Cancel the common factor $x-2$:
$\begin{align*}&=\lim_{x\to 2}\dfrac{x+3}{x+2}\end{align*}$
Now substitute $x=2$:
$\begin{align*}&\lim_{x\to 2}\dfrac{x^2+x-6}{x^2-4}\\&=\dfrac{2+3}{2+2}=\dfrac{5}{4}\end{align*}$
Method 2: Multiply by the conjugate
This method is for a special case. Use it when both of these are true:
- Direct substitution gives $\dfrac{0}{0}$.
- The numerator (or the denominator) is a sum or difference with a square root, such as $\sqrt{x+9}-3$.
It does not work for every function with a radical. For example, it does not remove a cube root. And when direct substitution works, as in Example 3, you do not need it.
The conjugate of $\sqrt{A}-B$ is $\sqrt{A}+B$. Their product has no square root:
$(\sqrt{A}-B)(\sqrt{A}+B)=A-B^2$.
So multiply the numerator and the denominator by the conjugate. This removes the square root from the numerator.
Example 6: Find $\displaystyle\lim_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}$.
Solution:
Direct substitution gives $0/0$. Multiply the numerator and the denominator by the conjugate of the numerator, $\sqrt{x+9}+3$. The numerator becomes a difference of squares:
$\begin{align*}&\lim_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}\\&=\lim_{x\to 0}\dfrac{(\sqrt{x+9})^2-3^2}{x(\sqrt{x+9}+3)}\end{align*}$
Simplify the numerator:
$\begin{align*}&=\lim_{x\to 0}\dfrac{(x+9)-9}{x(\sqrt{x+9}+3)}\\&=\lim_{x\to 0}\dfrac{x}{x(\sqrt{x+9}+3)}\end{align*}$
Cancel the common factor $x$:
$\begin{align*}&=\lim_{x\to 0}\dfrac{1}{\sqrt{x+9}+3}\end{align*}$
Now substitute $x=0$:
$\begin{align*}&\lim_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}\\&=\dfrac{1}{\sqrt{9}+3}=\dfrac{1}{6}\end{align*}$
Method 3: Combine fractions
Use this method when the numerator or the denominator has fractions in it. Write them as one fraction, using a common denominator. Then simplify.
Example 7: Find $\displaystyle\lim_{x\to 0}\dfrac{\dfrac{1}{x+2}-\dfrac{1}{2}}{x}$.
Solution:
Direct substitution gives $0/0$. Combine the two fractions in the numerator, using the common denominator $2(x+2)$:
$\begin{align*}&\lim_{x\to 0}\dfrac{\dfrac{1}{x+2}-\dfrac{1}{2}}{x}\\&=\lim_{x\to 0}\dfrac{\dfrac{2-(x+2)}{2(x+2)}}{x}\end{align*}$
Simplify the numerator. Dividing by $x$ puts $x$ in the denominator:
$\begin{align*}&=\lim_{x\to 0}\dfrac{-x}{2x(x+2)}\end{align*}$
Cancel the common factor $x$:
$\begin{align*}&=\lim_{x\to 0}\dfrac{-1}{2(x+2)}\end{align*}$
Now substitute $x=0$:
$\begin{align*}&\lim_{x\to 0}\dfrac{\dfrac{1}{x+2}-\dfrac{1}{2}}{x}\\&=\dfrac{-1}{2(0+2)}=-\dfrac{1}{4}\end{align*}$
Method 4: Expand and simplify
Use this method when the numerator has a power or a product that you can multiply out.
Example 8: Find $\displaystyle\lim_{h\to 0}\dfrac{(3+h)^2-9}{h}$.
Solution:
Here the variable is $h$ instead of $x$. The method is the same.
Direct substitution gives $0/0$.
Expand $(3+h)^2$:
$\begin{align*}&\lim_{h\to 0}\dfrac{(3+h)^2-9}{h}\\&=\lim_{h\to 0}\dfrac{9+6h+h^2-9}{h}\end{align*}$
Simplify the numerator, and factor out $h$:
$\begin{align*}&=\lim_{h\to 0}\dfrac{6h+h^2}{h}\\&=\lim_{h\to 0}\dfrac{h(6+h)}{h}\end{align*}$
Cancel the common factor $h$:
$\begin{align*}&=\lim_{h\to 0}(6+h)\end{align*}$
Let $h$ approach $0$. Put $0$ in for $h$:
$\begin{align*}&\lim_{h\to 0}\dfrac{(3+h)^2-9}{h}=6+0=6\end{align*}$
Limits like this one appear again when we study the derivative.
When direct substitution gives a nonzero number over 0
Example 9: Find $\displaystyle\lim_{x\to 2}\dfrac{x+1}{x-2}$.
Solution:
Direct substitution gives $\dfrac{3}{0}$. This is not an indeterminate form. The numerator approaches $3$, but the denominator approaches $0$. Dividing by a number very close to $0$ gives a very large number:
| $x$ | $\dfrac{x+1}{x-2}$ |
|---|---|
| 1.9 | −29 |
| 1.99 | −299 |
| 1.999 | −2999 |
| 2.001 | 3001 |
| 2.01 | 301 |
| 2.1 | 31 |
The values increase without bound from the right, and decrease without bound from the left. So the limit is not a finite number, and $\displaystyle\lim_{x\to 2}\dfrac{x+1}{x-2}$ does not exist. We study limits like this in infinite limits.
A strategy for computing limits
To find $\displaystyle\lim_{x\to a}f(x)$:
- Try direct substitution first.
- If you get a number, that number is the limit. (This is true for polynomials, rational functions, and their roots.)
- If you get $\dfrac{0}{0}$, change the form of the function. Factor and cancel, multiply by the conjugate, combine fractions, or expand. Then try direct substitution again.
- If you get a nonzero number over $0$, the limit is not a finite number. Check the one-sided limits. (See infinite limits.)