2.3 The Limit Laws

Limit laws let you break a complex limit into simpler pieces that you can evaluate individually. Once you know these rules, you can handle most limit problems in Calculus I by combining a few basic moves: direct substitution, algebraic simplification, and (when those fail) special techniques like the squeeze theorem.

Limit Laws and Techniques

Top images from around the web for Fundamental limit laws

• How Do You Compute the Limit of a Difference Quotient? – Math FAQ View original

  Is this image relevant?

• 

  The Product and Quotient Rules - Wisewire View original

  Is this image relevant?

• How Do You Evaluate The Limit Of A Difference Quotient? – Math FAQ View original

  Is this image relevant?

• How Do You Compute the Limit of a Difference Quotient? – Math FAQ View original

  Is this image relevant?

• The Product and Quotient Rules - Wisewire View original

  Is this image relevant?

Fundamental limit laws

The core idea is this: if $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, you can split up a complicated limit into parts. Each law below requires that the individual limits exist.

• Sum law: $\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$
• Difference law: $\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
• Product law: $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$
• Quotient law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$, provided $\lim_{x \to a} g(x) \neq 0$
• Power law: $\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n$ for any positive integer $n$
• Root law: $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}$ for any positive integer $n$ (and if $n$ is even, the limit inside must be positive)
• Constant multiple rule: $\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)$

There's also a composition law: if $\lim_{x \to a} g(x) = L$ and $f$ is continuous at $L$, then $\lim_{x \to a} f(g(x)) = f(L)$. This comes up whenever you have a function nested inside another, like $\lim_{x \to 0} \sin(x^2)$.

Limits of polynomial functions

Polynomials are continuous everywhere, so you can always evaluate their limits by direct substitution: just plug in the value of $a$.

For example: $\lim_{x \to 2} (3x^2 - 4x + 1) = 3(2)^2 - 4(2) + 1 = 12 - 8 + 1 = 5$

Rational functions (a polynomial divided by a polynomial) also allow direct substitution, as long as the denominator isn't zero at that point. When you plug in and get $\frac{0}{0}$, that's an indeterminate form, which signals that you need algebraic work before you can find the limit.

Simplifying complex limit expressions

When direct substitution gives you $\frac{0}{0}$, try these techniques:

Factoring and canceling: Factor the numerator and denominator, then cancel the common factor that's causing the zero.

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1}(x+1) = 2$

You can cancel $(x-1)$ because limits care about what happens near $x = 1$, not at $x = 1$.

Multiplying by the conjugate: When square roots are involved, multiply the numerator and denominator by the conjugate to eliminate the radical.

$\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \lim_{x \to 0} \frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}$

The conjugate trick works because $(A - B)(A + B) = A^2 - B^2$, which removes the square root from the numerator.

Application of the squeeze theorem

The squeeze theorem (also called the sandwich theorem) is your go-to when algebraic simplification won't work, especially with oscillating functions like sine and cosine.

The setup: if $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$ itself), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

You're "squeezing" $f(x)$ between two functions that both converge to the same value, so $f(x)$ has no choice but to converge there too.

Classic example: Show that $\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$.

1. You know $-1 \leq \sin\!\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$.
2. Multiply through by $x^2$ (which is non-negative): $-x^2 \leq x^2 \sin\!\left(\frac{1}{x}\right) \leq x^2$.
3. Both $\lim_{x \to 0}(-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$.
4. By the squeeze theorem, $\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$.

Specific limit strategies by function type

• Polynomial functions: Direct substitution always works.
• Rational functions: Try direct substitution first. If you get $\frac{0}{0}$, factor and cancel, or multiply by a conjugate.
• Trigonometric functions: Use the squeeze theorem or key identities. Two limits you should memorize: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.
• Exponential and logarithmic functions: These are continuous on their domains, so direct substitution typically works. Use properties of exponents and logarithms to simplify first when needed.

Function behavior near points

One-sided limits describe what happens as $x$ approaches $a$ from just one direction:

• From the left: $\lim_{x \to a^-} f(x)$
• From the right: $\lim_{x \to a^+} f(x)$

The two-sided limit $\lim_{x \to a} f(x)$ exists only if both one-sided limits exist and are equal. If they disagree, the two-sided limit does not exist, which indicates some kind of discontinuity at $x = a$.

Infinite limits occur when function values grow without bound. For example:

$\lim_{x \to 0^+} \frac{1}{x} = \infty \quad \text{and} \quad \lim_{x \to 0^-} \frac{1}{x} = -\infty$

Since the one-sided limits aren't equal (and aren't even finite), the two-sided limit $\lim_{x \to 0} \frac{1}{x}$ does not exist. The graph has a vertical asymptote at $x = 0$.

Continuity and limit definitions

A function $f$ is continuous at $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

This is exactly why direct substitution works for polynomials and other continuous functions: condition 3 says the limit equals the function value.

The epsilon-delta definition gives a rigorous way to prove a limit statement. It says $\lim_{x \to a} f(x) = L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - a| < \delta$, we have $|f(x) - L| < \epsilon$. You may or may not need to write epsilon-delta proofs in Calc I, but understanding the idea (making $f(x)$ as close to $L$ as you want by keeping $x$ close enough to $a$) helps the concept of a limit make sense.

The Intermediate Value Theorem (IVT) states that if $f$ is continuous on $[a, b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there exists some $c$ in $(a, b)$ with $f(c) = N$. This is often used to show that an equation has a solution: if $f(a)$ is negative and $f(b)$ is positive (or vice versa), then $f$ must cross zero somewhere between $a$ and $b$.