Limits and Algebraic Techniques: Direct Substitution, Cancellation, Conjugates, Vertical Asymptotes, and Piecewise Functions

Direct Substitution and Basic Limit Laws

If a limit point a does not cause any division by zero and the function is defined around a, then the limit is the same as the function value at a (direct substitution):
- If a constant c is involved, $\lim_{x\to a} c = c$ .
Basic limit laws (assuming the limits exist):
- $\lim{x\to a} [f(x) + g(x)] = \lim{x\to a} f(x) + \lim_{x\to a} g(x)$
- $\lim{x\to a} [f(x) - g(x)] = \lim{x\to a} f(x) - \lim_{x\to a} g(x)$
- $\lim{x\to a} [f(x) \cdot g(x)] = \lim{x\to a} f(x) \cdot \lim_{x\to a} g(x)$
- For quotients, provided $\lim{x\to a} g(x) \neq 0$ : $\lim{x\to a} \frac{f(x)}{g(x)} = \frac{\lim{x\to a} f(x)}{\lim{x\to a} g(x)}$
Constant multiples pull out of limits:
- $\lim{x\to a} [k \cdot f(x)] = k \cdot \lim{x\to a} f(x)$ where k is a constant.
If the inside limit does not involve the variable (the expression is constant with respect to x), the limit is that constant; e.g. a horizontal line y=c has limit c at any a.
Graphical intuition: a straight line like f(x) = x has limit a as x → a; a horizontal line g(x) = c has limit c as x → a.

When to use direct substitution first

Try plugging in a first. If you get a finite number (and no division by zero), that is the limit.
If you get 0/0 or an undefined form, you must simplify or manipulate the expression to resolve the indeterminacy.
If you get something like nonzero over zero, the limit tends to ±∞; classify with one-sided limits if needed.
In many problems, the goal is to algebraically cancel factors or rewrite the expression so the limit can be taken by direct substitution after simplification.

Example 1: f(x) = x and g(x) = constant

For any a, if we look at f(x) = x, then
- $\lim_{x\to a} x = a$ (graph: a straight line with value a at x=a).
For a constant function g(x) = c,
- $\lim_{x\to a} c = c$ (horizontal line at height c).

Example 2: Powers and repeated rules

If $\lim_{x\to a} f(x) = L$ and n is a positive integer,
- $\lim_{x\to a} [f(x)]^{n} = L^{n}$
For fractional powers (e.g., square roots) the same idea holds subject to domain restrictions and avoiding division by zero in the process.
In general, these follow from applying the limit laws to repeated multiplications and roots.

Example 3: The difference quotient (limit definition of the derivative)

Definition: $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Worked example:
- Evaluate $\lim_{h\to 0} \frac{(2+h)^{2} - 4}{h}$
- Expand: $(2+h)^2 = 4 + 4h + h^{2}$
- Numerator: $(4 + 4h + h^{2}) - 4 = 4h + h^{2}$
- Factor: $h(4 + h)$
- Cancel h: $4 + h$
- Take limit: $\lim_{h\to 0} (4 + h) = 4$
Note: This demonstrates going from an indeterminate form to a simple arithmetic limit by algebraic cancellation before substitution.

Example 4: Direct substitution in a simple rational expression

If substituting a into a rational expression yields a finite value (no division by zero), that value is the limit.
If substitution yields 0/0, proceed with simplification (factoring, canceling, etc.).
If substitution yields something like nonzero over zero, the limit is ±∞ and may require one-sided analysis.

Example 5: A conjugate (difference of squares) trick for square roots

When you have a limit with a square root in a quotient that causes 0/0, multiply by a conjugate to create a difference-of-squares pattern.
Example:
- Evaluate $\lim_{x\to 16} \frac{4 - \sqrt{x}}{16 - x}$
- Multiply numerator and denominator by the conjugate $4 + \sqrt{x}$ :
  $\frac{(4 - \sqrt{x})(4 + \sqrt{x})}{(16 - x)(4 + \sqrt{x})} = \frac{16 - x}{(16 - x)(4 + \sqrt{x})}$
- Cancel the common factor $16 - x$ (for $x
  e 16$):
  $\frac{1}{4 + \sqrt{x}}$
- Now substitute: $\lim_{x\to 16} \frac{1}{4 + \sqrt{x}} = \frac{1}{4 + 4} = \frac{1}{8}.$
Note: The transcript’s working suggested a slightly different intermediate form, but the correct cancellation yields the limit as $1/8$.

Example 6: A rational expression with a removable discontinuity

Consider $\lim_{x\to 4} \frac{x^{2} - 3x - 10}{x^{2} - 10x + 25}$
Factor numerator and denominator:
- Numerator: $x^{2} - 3x - 10 = (x - 5)(x + 2)$
- Denominator: $x^{2} - 10x + 25 = (x - 5)^{2}$
Cancel a common factor $(x - 5)$ (for $x \ne 5$):
- Simplified form: $\frac{x + 2}{x - 5}$
Now evaluate the limit as $x \to 4$ (not at the hole):
- $\lim_{x\to 4} \frac{x + 2}{x - 5} = \frac{4 + 2}{4 - 5} = \frac{6}{-1} = -6$.
If instead we consider $x \to 5$, the original function has a removable discontinuity at $x = 5$, since the simplified form has a hole there but the limit would be the value approached by the simplified expression (here $-3$). This illustrates the idea of removable discontinuities vs vertical asymptotes.

Example 7: A vertical asymptote and one-sided limits

A typical simple example: $\lim_{x \to 0} \frac{1}{x}$
- Right-hand limit: $\lim_{x \to 0^+} \frac{1}{x} = +\infty$
- Left-hand limit: $\lim_{x \to 0^-} \frac{1}{x} = -\infty$
- Since the one-sided limits do not agree, the two-sided limit does not exist.
A quick sign-check method (number line): identify where the function is undefined (division by zero) and where it equals zero (if ever). For 1/x, undefined at 0 and never zero; the sign of outputs changes across 0, giving opposite infinities on each side.

Example 8: One-sided limits and the two-sided limit at a piecewise boundary

When a function is piecewise, the limit at the boundary a depends on the left-hand branch (approach from the left) and the right-hand branch (approach from the right).
If the left-hand limit and the right-hand limit agree, the two-sided limit exists and equals that common value.
If they disagree, the two-sided limit does not exist.
Example structure (as discussed in class):
- If the left-hand approach yields 80 and the right-hand yields -1/2 at a breakpoint, the two-sided limit does not exist.

Example 9: Piecewise functions (brief preview)

In a piecewise setup, to evaluate lim_{x\to a} f(x), you take:
- left-hand limit: lim_{x\to a^-} f(x) using the left-hand piece,
- right-hand limit: lim_{x\to a^+} f(x) using the right-hand piece.
If the two are equal, the limit exists and equals that common value; otherwise, the limit does not exist.
The instructor provided a simple illustrated example: if the left-hand value is something like 45 and the right-hand value is something like 0.5, the limit does not exist.

Class logistics and administration (from the lecture)

First homework is due this Wednesday at midnight (the instructor noted they started it).
The first quiz is planned for Friday; the instructor often schedules quizzes on Fridays and office hours before the quiz.
Office hours timing: usually earlier in the week; Friday office hours might be skipped if many students want another day.
If you need the link re-sent for assignments, the instructor is okay with resending; the online schedule will be updated.
The focus of today’s class: algebraic limit techniques, particularly limits not involving the graph, and preparation for upcoming topics (notably algebraic limits and absolute values).
The class also referenced the plan to cover one more topic: piecewise functions, then proceed to more advanced limit techniques in subsequent sessions.

Key takeaways to remember for the exam

When facing a limit, always try direct substitution first:
- If it works, that’s the limit.
- If you get 0/0, look for algebraic simplifications that cancel factors or rationalize expressions.
- If you get nonzero over zero, the limit is ±∞; analyze one-sided limits as needed.
Use limit laws to break apart sums, differences, products, and quotients, and pull out constants when possible.
For expressions with square roots, consider conjugates to create a difference-of-squares structure to cancel problematic factors.
For piecewise functions, determine the left-hand and right-hand limits at a boundary to decide whether the two-sided limit exists.
Be aware of potential removable discontinuities (holes) versus genuine vertical asymptotes (unbounded behavior).