Topic 1.6 Notes – Determining Limits Using Algebraic Manipulation

What Algebraic Manipulation Does to a Limit

Start every limit the same way. Plug the value in.

• If you get a real number, that number is the limit.
• If you get $\frac{0}{0}$, that’s an indeterminate form. The limit might exist, but you need to rewrite the expression.
• If you get something like nonzero over 0, that usually signals a vertical asymptote (infinite limit or DNE).

The key idea behind everything in this section:

If two expressions are equal everywhere near $x=c$ (even if one is undefined at $c$), they have the same limit as $x \to c$.

So we manipulate the algebra to create an equivalent expression that’s easier to evaluate.

Factoring and Canceling Common Factors

This is the most common situation. Usually you’ll see a rational function where substitution gives $\frac{0}{0}$.

Example structure:

$$\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$$

Plug in 4 → $\frac{0}{0}$. So factor:

$$x^2 - 16 = (x-4)(x+4)$$

Now:

$$\frac{(x-4)(x+4)}{x-4}$$

Cancel the common factor (for $x \ne 4$):

$$x+4$$

Now evaluate the limit:

$$x+4 \to 8$$

So the limit is 8.

What’s happening graphically?

When you cancel the factor, the function behaves like the line $y = x+4$ everywhere except at $x=4$.

That open circle at $(4,8)$ is a hole. The function isn’t defined at 4, but it approaches 8 from both sides. That’s why the limit exists.

Common factoring patterns

• Difference of squares: $a^2 - b^2 = (a-b)(a+b)$
• Quadratics
• Factoring out a GCF
• Polynomial division if degree of numerator ≥ denominator

Students often try canceling terms instead of factors. You can cancel $(x-4)$, but not just the 4’s inside separate terms.

Rationalizing with Conjugates

If radicals cause the $\frac{0}{0}$, factoring won’t help. That’s when you use a conjugate.

The conjugate of $a + b$ is $a - b$.

Example:

$$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$$

Plug in 9 → $\frac{0}{0}$.

Multiply by the conjugate:

$$\frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3}$$

Use difference of squares:

$$(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9$$

So the expression becomes:

$$\frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$$

Cancel $x-9$:

$$\frac{1}{\sqrt{x} + 3}$$

Now substitute:

$$\frac{1}{3+3} = \frac{1}{6}$$

Done.

This method shows up constantly on no-calculator sections. The biggest mistake is forgetting to multiply the entire fraction by the conjugate.

Trig Limits and Equivalent Forms

Certain trig limits are foundational:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

$$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$

Also, sine and cosine are continuous:

$$\lim_{x \to c} \sin x = \sin c$$

Using the sine limit with constants

Example:

$$\lim_{x \to 0} \frac{\sin(5x)}{x}$$

Rewrite:

$$\frac{\sin(5x)}{5x} \cdot 5$$

Now apply the rule → result is 5.

If you forget to adjust for that constant, you’ll lose points fast on MCQs.

Ratios like $\frac{\sin(7x)}{\sin(2x)}$

Rewrite each to match the sine-over-angle pattern:

$$\frac{\sin(7x)}{7x} \cdot \frac{2x}{\sin(2x)} \cdot \frac{7}{2}$$

Each sine-over-angle → 1, so limit is $\frac{7}{2}$.

The Squeeze Theorem

This is used when algebra alone doesn’t simplify things, often with oscillating functions.

If:

$$g(x) \le f(x) \le h(x)$$

and both outer functions approach the same limit $L$, then $f(x)$ also approaches $L$.

Classic example:

$$x \sin\left(\frac{1}{x}\right)$$

Since

$$-1 \le \sin\left(\frac{1}{x}\right) \le 1$$

Multiply everything by $x$:

$$-x \le x\sin\left(\frac{1}{x}\right) \le x$$

As $x \to 0$, both outer expressions go to 0, so the middle must also go to 0.

Graph of $y = x\sin(1/x)$ bounded by $y = x$ and $y = -x$

The graph shows the function oscillating between the lines $y = x$ and $y = -x$, with the oscillations shrinking toward 0 as $x \to 0$.

On FRQs, you often need to state the inequality explicitly to earn full credit.