Indeterminate Forms

• An expression is in an indeterminate form at $x = c$ if $\frac{f(c)}{g(c)}$ results in $\frac{0}{0}$.

• Example: If $\text{let } \theta \text{ be an angle between } 0 \text{ and } 2\pi$ and we plug in $\theta = 0$, then we have (\frac{\sin(0)}{0} = \frac{0}{0}), which is indeterminate.

Solving Limits of Indeterminate Forms

• To find the limit as $x$ approaches $c$ of $\frac{f}{g}$, the first step is algebraic reduction of (\frac{f}{g}) to a simpler function (h(x)).

• Note that sometimes step one cannot be performed easily.

• Example Demonstration: Instead of directly reducing $\frac{\sin(\theta)}{\theta}$, we relate $\sin^2(\theta)$ to $1 - \cos^2(\theta)$ which is derived from the Pythagorean theorem.

The Squeeze Theorem

• The Squeeze Theorem (T) states that if you can find two functions $g(x)$ and $h(x)$ such that:
  • (h(x) \leq f(x) \leq g(x)) for $x$ in some vicinity of $c$ and
  • (\lim{x \to c} g(x) = \lim{x \to c} h(x) = L)
• Then: (\lim_{x \to c} f(x) = L).

Conditions for the Squeeze Theorem

1. Bounding functions: The function $f(x)$ must be bounded between $g(x)$ and $h(x)$, where:

   • $g(x)$ is greater than or equal to $f(x)$.
   • $h(x)$ is less than or equal to $f(x)$.

2. Same limit: Both $g(x)$ and $h(x)$ must have the same limit as $x$ approaches $c$.

Application of Squeeze Theorem to $\frac{\sin(\theta)}{\theta}$

• Step 1: Let $f(\theta) = \frac{\sin(\theta)}{\theta}$. We need to find functions $g(\theta)$ and $h(\theta)$.

• Step 2: Identify $g(\theta)$ and $h(\theta)$ such that $\frac{\sin(\theta)}{\theta}$ is squeezed between them.

Geometric Interpretation

• Using a unit circle: For small angles, as $\theta$ approaches $0$, we look at:

  • Area A (Triangle): Area of the yellow triangle = (\frac{1}{2} \times 1 \times \sin(\theta) = \frac{1}{2} \sin(\theta)).

  • Area B (Sector): The area of the sector of the circle is relative to $\theta$: (\frac{\theta}{2\pi} \text{ (of unit circle)}), hence Area B = (\frac{\theta}{2\pi} \times \pi = \frac{\theta}{2}).

  • Conclusion on Areas:

    • Area A < Area B (Triangle covers less than the sector).

• Thus, for $\sin(\theta)$ when $ heta$ approaches $0$:

  • (\frac{1}{2} \sin(\theta) \leq \frac{\theta}{2} \leq \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \implies \sin(\theta) < \theta < \tan(\theta)).

Limit Calculation

• Dividing the inequalities by $\sin(\theta)$ gives:

  • (1 < \frac{\theta}{\sin(\theta)} < \frac{1}{\cos(\theta)}).

• Applying the Squeeze Theorem:

  • As $\theta \to 0$, both sides converge to 1, thus:
  • (\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1).

Additional Limit - Cosine Case

• Discussed limit: For (\frac{1 - \cos(\theta)}{\theta}) as $\theta \to 0$ also leads to an indeterminate form. This topic is suggested to be explored independently as it relates to trigonometric identities, especially noting that (\sin^2(\theta) + \cos^2(\theta) = 1).