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Indeterminate Forms

  • An expression is in an indeterminate form at $x = c$ if f(c)g(c)\frac{f(c)}{g(c)} results in 00\frac{0}{0}.

  • Example: If let θ be an angle between 0 and 2π\text{let } \theta \text{ be an angle between } 0 \text{ and } 2\pi and we plug in θ=0\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 fg\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 sin⁡(θ)θ\frac{\sin(\theta)}{\theta}, we relate sin⁡2(θ)\sin^2(\theta) to 1−cos⁡2(θ)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 sin⁡(θ)θ\frac{\sin(\theta)}{\theta}

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

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

Geometric Interpretation

  • Using a unit circle: For small angles, as θ\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 θ\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⁡(θ)\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⁡(θ)\sin(\theta) gives:

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

  • Applying the Squeeze Theorem:

    • As θ→0\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 θ→0\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).