Evaluating Limits Algebraically

Welcome, everyone! Now that we understand what a limit is visually and numerically, it’s time to learn how to find them efficiently using algebra. While graphs and tables are great for intuition, algebraic techniques are fast, precise, and powerful.

Part 1: The Golden Rule – Just Plug It In!

The first and simplest technique for evaluating a limit is direct substitution.

The Rule: If a function $f(x)$ is continuous at $x=a$, then:

$$\lim_{x\to a}f(x)=f(a)$$

When does this work? For most of the functions you’ve known for years! Polynomials, radicals, exponentials, sines, and cosines are continuous on their domains.

Example 1:

$$\lim_{x\to3}(2x^{2}-5x+1)$$

This is a polynomial, so it’s continuous everywhere. Just plug in $x=3$:

$$2(3)^{2}-5(3)+1=4$$

Example 2:

$$\lim_{x\to\pi}sin(x)$$

The sine function is continuous everywhere. Just plug in $$x=\pi$$:

$$sin(\pi)=0$$

Part 2: When Plugging In Fails – The Indeterminate Form $$\frac{0}{0}$$​

Often, especially with rational functions (fractions), direct substitution leads to $$\frac{0}{0}$$​. This is not an answer! It’s an indeterminate form, which means the limit could be 0, a finite number, infinity, or might not exist. It tells us we need to do more work to uncover the limit’s true value.

Our main strategy is to algebraically manipulate the function into a new form where direct substitution will work.

Technique 1: Factoring and Canceling

This is the go-to method for rational functions where both the numerator and denominator go to zero.

Example 3:

$$\lim_{x\to2}\frac{x^{2}-4}{x-2}$$​

1. Try Direct Substitution: $$\frac{(2)^2-4}{x-2}=\frac{0}{0}$$→ Indeterminate.

2. Factor and Simplify: Notice the numerator is a difference of squares.

   $$\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}$$​

   For all x≠2, we can cancel the $(x-2)$ terms.

   $$=x+2$$

3. Take the Limit of the Simplified Form: Now we have $$\lim_{x\to2}(x+2)$$. This is a continuous function, so we can plug in $x=2$.

   =2+2=4

   So,$$\lim_{x\to2}\frac{x^{2}-4}{x-2}=4$$.

Why is this valid? Remember, the limit as $$x\rightarrow2$$ doesn’t care about the value at 2, only the values around it. For all x values approaching 2 (but not equal to 2), the original function and the simplified function x+2 are identical.

Technique 2: Rationalizing the Numerator or Denominator

This technique is essential when dealing with radicals (square roots).

Example 4:

$$\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}$$​

1. Try Direct Substitution: $$\frac{\sqrt{0+4}-2}{0}=\frac{2-2}{0}=\frac{0}{0}$$ → Indeterminate.

2. Rationalize the Numerator: Multiply the top and bottom by the conjugate of the numerator. The conjugate of $$\sqrt{x+4}-2$$ is $$\sqrt{x+4}+2$$.

   $$\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$$​

3. Simplify: 

   $$\frac{(\sqrt{x+4})^{2}-2^{2}}{x(\sqrt{x+4}+2)}=\frac{x+4-4}{x(\sqrt{x+4}+2)}=\frac{x}{x(\sqrt{x+4}+2)}$$​

4. Cancel the Common Factor: Cancel the $x$ from the numerator and denominator.

   $$\frac{1}{(\sqrt{x+4}+2)}$$​

5. Take the Limit of the Simplified Form: Now use direct substitution with x=0.

   $$\lim_{x\to0}\frac{1}{(\sqrt{x+4}+2)}=\frac{1}{\sqrt{4}+2}=\frac{1}{2+2}=\frac{1}{4}$$​

Part 3: Limits at Infinity – The End Behavior of Functions

We also use limits to describe what happens to a function as $x$ becomes very large (x→∞) or very small (x→−∞).

The key technique for rational functions $$\frac{f(x)}{g(x)}$$ is to divide every term by the highest power of $x$ in the denominator.

Example 5:

$$\lim_{x\to\infty}\frac{3x^{2}-2x+5}{2x^{2}+7}$$​

1. The highest power of $x$ in the denominator is $$x^{2}$$. Divide every term in the numerator and denominator by $$x^{2}$$.$$=\lim_{x\to\infty}\frac{\frac{3x^2}{x^2}-\frac{2x}{x^2}+\frac{5}{x^2}}{\frac{2x^2}{x^2}+\frac{7}{x^2}}=\lim_{x\to\infty}\frac{3-\frac{2}{x}+\frac{5}{x^2}}{2+\frac{7}{x^2}}$$

2. now, ​here is important $$\lim_{x\to\infty}\frac{1}{x}=0$$

3. Evaluate the Limit: As x→∞, terms like $$\frac{2}{x},\frac{5}{x^{2}},\frac{7}{x^{2}}$$ all approach 0.

   $$=\frac{3-0+0}{2+0}=\frac{3}{2}$$​

General Rule for Rational Functions at Infinity:
Compare the degrees of the numerator ($N$) and denominator ($D$).

• If Degree(N) < Degree(D), the limit is 0.

• If Degree(N) = Degree(D), the limit is the ratio of the leading coefficients.

• If Degree(N) > Degree(D), the limit is ∞ or −∞.

Part 4: The Squeeze Theorem

Sometimes, a function is too complicated to handle directly. The Squeeze Theorem allows us to trap it between two simpler functions.

The Squeeze Theorem:
If g(x)≤f(x)≤h(x) for all $x$ near $a$ (except possibly at a), and

$$\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$$

then it follows that:

$$\lim_{x\to a}f(x)=L$$

Basic Example 6:

lim⁡x→0sin⁡xxx→0lim​xsinx​

You can’t factor or rationalize this. Using geometry, we can prove that for $x$ near 0,$$\cos(x)\leq\frac{sin(x)}{x}\leq1.$$
We know $$\lim_{x\to0}\cos(x)=1$$ and \lim_{x\to0}1=1.
Since $$\frac{\sim(x)}{x}$$ is “squeezed” between two functions that both approach 1, it must also approach 1.

$$\lim_{x\to0}\frac{sim(x)}{x}=1$$