Algebraic Rules and Techniques for Limits and Continuity

Infinite Limits

Sometimes, as $x$ approaches a value $a$, the function does not settle on a finite number but instead grows without bound. In such cases, we describe the limit as an infinite limit.

Limits Involving Infinity

If one part of your expression approaches infinity while the other approaches a constant $B$, the behavior of the entire expression depends on the interaction between the two:

• Addition: If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = B$, then $\lim_{x \to a} (f(x) + g(x)) = \infty$.
• Division: If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = B$, then $\lim_{x \to a} \frac{g(x)}{f(x)} = 0$.
• Multiplication: If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = B$, the result is $\infty$ if $B > 0$, and $-\infty$ if $B < 0$.

Polynomial Behavior at Infinity

When evaluating the limit as $x$ approaches positive or negative infinity ($\pm\infty$), we look at the power $n$ of the term $x^n$:

1. $\lim_{x \to \infty} x^n = \infty$ for $n \in \mathbb{N}$.
2. $\lim_{x \to -\infty} x^n = \infty$ if $n$ is even, and $-\infty$ if $n$ is odd.
3. $\lim_{x \to \pm\infty} \frac{1}{x^n} = 0$.

Techniques for Evaluating Complex Limits

When direct substitution leads to an "indeterminate form" (such as $\frac{0}{0}$ or $\infty - \infty$), you must use algebraic techniques to simplify the expression before re-evaluating the limit.

Rationalizing the Denominator (or Numerator)

If an expression involves square roots and results in an indeterminate form, you can often resolve it by multiplying by the conjugate.

Example: Evaluating $\lim_{x \to 1} \frac{x-1}{\sqrt{x}-1}$ To solve this, we multiply the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{x} + 1)$:

$$\begin{aligned} \lim_{x \to 1} \frac{x-1}{\sqrt{x}-1} &= \lim_{x \to 1} \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \\ &= \lim_{x \to 1} \frac{(x-1)(\sqrt{x}+1)}{x-1} \\ &= \lim_{x \to 1} (\sqrt{x}+1) \\ &= \sqrt{1}+1 = 2 \end{aligned}$$

Limits of Rational Functions at Infinity

To evaluate $\lim_{x \to \infty} \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials, the most effective method is to divide both the numerator and the denominator by the highest power of $x$ found in the denominator.

This technique relies on the fact that $\lim_{x \to \infty} \frac{1}{x^n} = 0$. Depending on the degrees of the polynomials, three outcomes are possible:

1. Degree of Numerator < Degree of Denominator: The limit is $0$.
2. Degree of Numerator = Degree of Denominator: The limit is the ratio of the leading coefficients.
3. Degree of Numerator > Degree of Denominator: The limit approaches $\pm\infty$.

Example: Evaluating $\lim_{x \to \infty} \frac{6x^2-x-7}{3x^2-3x+1}$ Divide every term by $x^2$ (the highest power in the denominator):

$$\begin{aligned} \lim_{x \to \infty} \frac{6x^2-x-7}{3x^2-3x+1} &= \lim_{x \to \infty} \frac{6 - \frac{1}{x} - \frac{7}{x^2}}{3 - \frac{3}{x} + \frac{1}{x^2}} \\ &= \frac{6 - 0 - 0}{3 - 0 + 0} \\ &= \frac{6}{3} = 2 \end{aligned}$$

Summary of Continuity

Finally, limits are the foundation of continuity. A function $f(x)$ is considered continuous at a point $a$ if and only if three conditions are met:

1. $f(a)$ is defined (the point exists in the domain).
2. $\lim_{x \to a} f(x)$ exists (the left-hand and right-hand limits are equal).
3. $\lim_{x \to a} f(x) = f(a)$ (the limit matches the function's value).

If any of these conditions fail, the function is discontinuous at that point.