What you’ll learn:

• How to simplify indeterminate forms with algebraic techniques: factoring, rationalizing, and common denominators
• Analyze limits with absolute value functions

Indeterminate forms

In the previous section, we found limits by substituting the value of $x$ into the function. Sometimes, though, direct substitution produces an undefined or unhelpful expression like $\frac{0}{0}$. This is called an indeterminate form. It doesn’t mean the limit doesn’t exist - it means we can’t determine the limit from substitution alone.

There are only 7 indeterminate forms:

Indeterminate forms:

1. $0/0$
2. $\pm \infty /\pm \infty$
3. $0\times \infty$
4. $\infty -\infty$
5. $0^0$
6. $1^{\infty }$
7. ${\infty }^0$

To resolve these, you’ll usually rewrite the expression using algebraic techniques - such as factoring, rationalizing, or combining terms with common denominators - and then try substitution again.

Factoring

Many limits simplify after factoring. The main idea is to rewrite the expression so you can cancel the factor that creates the “problem area,” turning an indeterminate form into something you can evaluate.

Example

Evaluate

$$\underset{x\to -2}{\lim }\frac{x^2-4}{3x^3+24}$$

Solution

Start with direct substitution:

$$\begin{array}{rl}& \underset{x\to -2}{\lim }\frac{x^2-4}{3x^3+24}\\ & =\frac{(-2{)}^2-4}{3(-2{)}^3+24}\\ & =\frac{0}{0}\end{array}$$

This is an indeterminate form.

Now factor the numerator and denominator.

• Numerator: difference of squares
• Denominator: factor out the GCF, then factor the remaining sum of cubes

$$\underset{x\to -2}{\lim }\frac{(x+2)(x-2)}{3(x^3+8)}$$

The expression $(x^3+8)$ matches the sum of cubes pattern $(a^3+b^3)$. To use the formula, identify $a$ and $b$:

Let $a^3=x^3$ so $a=x$. Let $b^3=8$ so $b=2$.

Factoring gives:

Factoring: $(x^3+8)=(x+2)(x^2-2x+4)$

Substitute that into the limit and cancel the common factor:

$$\underset{x\to -2}{\lim }\left(\frac{(x-2)\cancel{(x+2)}}{3\cancel{(x+2)}(x^2-2x+4)}\right)$$

$$\begin{array}{rl}=\underset{x\to -2}{\lim }& \left(\frac{x-2}{3(x^2-2x+4)}\right)\\ & =\frac{-4}{3(12)}\\ & =\boxed{-\frac{1}{9}}\end{array}$$

AP tip:

Keep the limit notation in front of the expression at every step until you actually substitute the value into $x$. Dropping the limit notation too early can make it look like you’re claiming the limit equals the function value, which isn’t automatically true.

Rationalization with conjugates

This technique is especially useful when square roots lead to an indeterminate form after substitution. The idea is to multiply by a conjugate so the square root expression simplifies using a difference of squares.

Conjugate:
A binomial expression with the same two terms but the opposite sign between them. For example, the conjugate of$(a+b)$ is $(a-b)$.

For example, the conjugate of $(\sqrt{x-2}+5)$ is $(\sqrt{x-2}-5)$. Multiplying them gives a difference of squares:

$$(\sqrt{x-2}+5)(\sqrt{x-2}-5)$$

$$=(\sqrt{(x-2{)}^2}-5^2)$$

$$=((x-2)-25)$$

$$=(x-27)$$

AP tip:

Since the goal is to cancel the factor that becomes $0$, expand only the conjugate product. Keep other parts factored so it’s easier to see what cancels.

Example

Evaluate

$$\underset{x\to 4}{\lim }\frac{\sqrt{x}-2}{4-x}$$

Solution

Here is an alternative method for the same problem that uses factoring:

Combining with common denominators

When you see a “fraction over a fraction,” it often helps to combine terms using a common denominator first.

Example

Evaluate

$$\underset{x\to 2}{\lim }\frac{\frac{1}{x}-\frac{1}{2}}{x-2}$$

Solution

Direct substitution gives the indeterminate form $\frac{0}{0}$. To combine the two fractions in the numerator, use the common denominator $2x$.

$$\underset{x\to 2}{\lim }\frac{\frac{1}{x}\cdot \frac{2}{2}-\frac{1}{2}\cdot \frac{x}{x}}{x-2}$$

$$=\underset{x\to 2}{\lim }\frac{\frac{2}{2x}-\frac{x}{2x}}{x-2}$$

$$=\underset{x\to 2}{\lim }\frac{(\frac{2-x}{2x})}{x-2}$$

Now rewrite the complex fraction as multiplication by the reciprocal:

$$\underset{x\to 2}{\lim }(\frac{2-x}{2x}\times \frac{1}{x-2})$$

Rewrite $(2-x)$ as $-(x-2)$ so the factor cancels:

$$\underset{x\to 2}{\lim }(\frac{-\cancel{(x-2)}}{2x}\times \frac{1}{\cancel{(x-2)}})$$

$$=\underset{x\to 2}{\lim }\frac{-1}{2x}$$

$$=\boxed{-\frac{1}{4}}$$

Absolute value functions

To find $\underset{x\to 0}{\lim }|x|$, direct substitution works and the limit is $0$. But absolute values can create indeterminate forms when the expression inside the bars becomes $0$.

For example, suppose

$$f(x)=\frac{x^2-3x+2}{|x-1|}$$

and you want to find $\underset{x\to 1}{\lim }f(x)$. Direct substitution gives $\frac{0}{0}$.

To resolve this, rewrite the absolute value as a piecewise function. Then evaluate the one-sided limits separately.

Recall the definition of absolute value:

$$|x|=\{\begin{array}{ll}x& \text{if }x\ge 0\\ & \\ -x& \text{if }x<0\end{array}$$

To rewrite an absolute value expression as a piecewise function, first find the breakpoint(s) - the value(s) of $x$ that make the expression inside the bars equal to $0$.

Rewrite $g(x)=|x-1|$ as a piecewise function.

The expression inside the bars is $x-1$, which equals $0$ when $x=1$. To decide which piece applies on each side of $1$, test a point in each region.

When $x>1$:
Choose a point greater than 1, like$x=2,$ and plug into $(x-1)$. Since $2-1=1$ (positive), the piece of the function defined for all $x>1$ is:

$$+(x-1)$$

When $x<1$:
Choose a point less than 1, like$x=0,$ and plug into $(x-1)$. Since $0-1=-1$ (negative), the piece defined for all $x<1$ is:

$$-(x-1)$$

At $x=1$:
Both pieces evaluate to$0$. By convention, the equality is usually included with the $x>$ side unless the problem gives a reason to do otherwise.

So $g(x)$ written as a piecewise function is:

$$g(x)=\{\begin{array}{ll}x-1& \text{if }x\ge 1\\ & \\ -(x-1)& \text{if }x<1\end{array}$$

And $f(x)$ written in piecewise form is

$$f(x)=\{\begin{array}{ll}\frac{x^2-3x+2}{x-1}& \text{if }x\ge 1\\ & \\ \frac{x^2-3x+2}{-(x-1)}& \text{if }x<1\end{array}$$

To find $\underset{x\to 1}{\lim }f(x)$, evaluate the one-sided limits.

• Left:

$$\begin{array}{rl}& \underset{x\to 1^{-}}{\lim }\frac{x^2-3x+2}{x-1}\\ & =\underset{x\to 1^{-}}{\lim }\frac{(x-2)\cancel{(x-1)}}{\cancel{(x-1)}}\\ & =\underset{x\to 1^{-}}{\lim }(x-2)\\ & =\boxed{-1}\end{array}$$

• Right:

$$\begin{array}{rl}& \underset{x\to 1^{+}}{\lim }\frac{x^2-3x+2}{-(x-1)}\\ & =\underset{x\to 1^{+}}{\lim }\frac{(x-2)\cancel{(x-1)}}{-\cancel{(x-1)}}\\ & =\underset{x\to 1^{+}}{\lim }\frac{(x-2)}{-1}\\ & =\boxed{1}\end{array}$$

Since the one-sided limits don’t match, $\underset{x\to 1}{\lim }f(x)$ does not exist.

Challenge problems

1. Evaluate

$$\underset{x\to 1}{\lim }\frac{x^{1/3}-1}{x-1}$$

Solution

2. Evaluate

$$\underset{x\to 0}{\lim }\frac{|x-1|-|x+1|}{x}$$

Solution