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 the limits by simply substituting the value of $x$ into the function. But sometimes direct substitution leads to a messy or undefined expression like $\frac{0}{0}$. This is called an indeterminate form, which means we can’t determine the limit just yet, not necessarily that it doesn’t exist! 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, we’ll have to algebraically manipulate the expression using techniques like factoring, rationalizing, or combining with common denominators before retrying direct substitution.

1. Factoring

Many limits can be simplified with factoring. The primary goal is to rewrite the expression and cancel out the “problem area,” turning the indeterminate form into a solvable one.

Example

Evaluate

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

Solution

First, try 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.

Factoring the difference of squares in the numerator and the GCF in the denominator,

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

$(x^3+8)$ is similar in form to the sum of cubes $(a^3+b^3)$. As an organization tip, define $a$ and $b$ to plug into the factoring formula.

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

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

Then

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

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

AP tip:

Make sure every step has the limit notation written in front of $f(x)$, up until $a$ is directly substituted into $x$! Or else it will mean that the limit equals the function $f(x)$, which isn’t true. Graders will take points off for this mistake.

2. Rationalization with conjugates

This technique is useful for limits with square roots if direct substitution results in an indeterminate form. Look for the expression with a square root and multiply by its conjugate.

Conjugate:
A binomial expression that has the same two terms but the opposite sign in between e.g. the conjugate of $(a+b)$ is $(a-b)$.

For example, the conjugate of $(\sqrt{x-2}+5)$ is $(\sqrt{x-2}-5)$. Multiplied together 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 out the portion that results in $0$, it’s best to expand only the expression with the conjugate pair and leave everything else in factored form.

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:

3. Combining with common denominators

When there are complex expressions involving fractions on fractions, combine them.

Example

Evaluate

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

Solution

Direct substitution results in the indeterminate form $\frac{0}{0}$. To combine the two rational expressions in the numerator, both must have the common denominator of $2x$.

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

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

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

Simplify the complex fraction by rewriting division as multiplication:

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

Rearrange $(2-x)$ into $-(x-2)$ so it can be canceled out:

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

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

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

Absolute value functions

To find $\underset{x\to 0}{\lim }|x|$, direct substitution works well and the limit is $0$. But suppose

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

and you wanted to find $\underset{x\to 1}{\lim }f(x)$. Direct substitution in this case results in $\frac{0}{0}$.

To resolve this indeterminate form, the absolute value function must be rewritten in piecewise form. One-sided limits are then evaluated separately.

Recall that the absolute value function is defined as

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

To turn any absolute value function into a piecewise function, first identify the breakpoint(s), which are where the expression inside the bars $=0$.

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

The expression between the absolute value bars equals 0 when $x=1$. To determine how the function behaves on either side, test points in each region:

1. 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:

$$+(x-1)$$

2. When $x<1$:
   A point that is less than 1, like $x=0,$ results in $0-1=-1$ (negative). So the piece defined for all $x<1$ is:

$$-(x-1)$$

3. At $x=1$:
   Both pieces evaluate to $0$. By convention, the equal sign is put under the sign where $x>$ unless there’s a reason not to do so (e.g. a given restriction in a problem).

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 separately:

• 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 agree, $\underset{x\to 1}{\lim }f(x)$ 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

Key points

• When substitution gives an indeterminate form, this does not mean the limit does not exist. Try to factor, rationalize, simplify, or combine techniques and re-evaluate.
• For limits with absolute values, rewrite as a piecewise function and evaluate the one-sided limits separately.