Limits (intro) Knowledge-Based Study

Study notes

1. What is a Limit?

   In calculus, the concept of a limit describes the behavior of a function near a specific point, rather than at that point itself.

   Suppose we have a function $f(x)$. When we say that the limit of $f(x)$ as $x$ approaches a number $c$ is equal to $L$, we mean that the value of $f(x)$ gets arbitrarily close to $L$ as $x$ gets sufficiently close to $c$ from both sides, while $x \neq c$.

   Mathematically, we write this as:

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

   Key points to remember:

   • A limit examines values of $x$ near $c$, not at $x = c$.
   • The actual function value $f(c)$ does not need to exist for $\lim_{x \to c} f(x)$ to exist.

   Approaching from Left and Right

   To understand whether a two-sided limit exists, we inspect the behavior of the function from each direction independently:

   • Left-hand limit: The value that $f(x)$ approaches as $x$ approaches $c$ from values strictly less than $c$, written as $\lim_{x \to c^-} f(x)$.
   • Right-hand limit: The value that $f(x)$ approaches as $x$ approaches $c$ from values strictly greater than $c$, written as $\lim_{x \to c^+} f(x)$.

   Existence Criterion

   The general limit $\lim_{x \to c} f(x)$ exists and equals $L$ if and only if both one-sided limits exist and are equal to each other:

   $\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L \text{ and } \lim_{x \to c^+} f(x) = L$

   If the left-hand and right-hand limits differ, or if either fails to exist, the two-sided limit does not exist (DNE).

   Direct Substitution

   For elementary functions such as polynomials and rational functions with non-zero denominators, the limit at $x = c$ is simply equal to the function evaluated at that point:

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

   Worked Example

   Evaluate $\lim_{x \to 2} (3x^2 - 4x + 5)$:

   Substitute $x = 2$ directly into the polynomial:

   $3(2)^2 - 4(2) + 5 = 3(4) - 8 + 5 = 12 - 8 + 5 = 9$

   Therefore, $\lim_{x \to 2} (3x^2 - 4x + 5) = 9$.

   The Indeterminate Form $\frac{0}{0}$

   When direct substitution produces the expression $\frac{0}{0}$, direct evaluation fails. This indicates that the numerator and denominator share a common factor that can often be simplified algebraically.

   Example: Factoring Technique

   Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$:

   1. Attempt direct substitution: $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$ (indeterminate).
   2. Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$.
   3. Simplify the fraction for all $x \neq 3$:
      $\frac{(x - 3)(x + 3)}{x - 3} = x + 3$
   4. Evaluate the simplified limit:
      $\lim_{x \to 3} (x + 3) = 3 + 3 = 6$

   Even though the original function is undefined at $x = 3$, the limit exists and equals $6$.