Squeeze theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is a fundamental concept in calculus that helps determine the limit of a function. It states that if you have three functions f(x)f(x), g(x)g(x), and h(x)h(x) defined in a neighborhood around a point aa, and if f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) for all xx in that neighborhood (except possibly at aa), and if the limits of f(x)f(x) and h(x)h(x) as xx approaches aa are both equal to LL, then the limit of g(x)g(x) as xx approaches aa is also equal to LL.

In summary, it "squeezes" the function g(x)g(x) between two other functions whose limits are known to be the same.

Part 1: Squeeze theorem intro

The squeeze (or sandwich) theorem states that if f(x)≤g(x)≤h(x) for all numbers, and at some point x=k we have f(k)=h(k), then g(k) must also be equal to them. We can use the theorem to find tricky limits like sin(x)/x at x=0, by "squeezing" sin(x)/x between two nicer functions and ​using them to find the limit at x=0.

Certainly! Here are the key points to focus on when studying the Squeeze Theorem:

  1. Definition: The Squeeze Theorem (also known as the Sandwich Theorem) states that if a function f(x)f(x) is squeezed between two other functions g(x)g(x) and h(x)h(x), and if g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) for all xx in an interval (except possibly at a point), then if limxag(x)=limxah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, it follows that limxaf(x)=L\lim_{x \to a} f(x) = L.

  2. Conditions:

    • The functions gg and hh must both converge to the same limit LL as xx approaches aa.
    • The function f(x)f(x) must be bounded by g(x)g(x) and h(x)h(x) in the specified interval.
  3. Use Cases: This theorem is particularly useful for finding limits of functions that are difficult to evaluate directly using algebra or L'Hôpital's rule.

  4. Visual Representation: A graphical illustration shows how f(x)f(x) is constrained between g(x)g(x) and h(x)h(x), reinforcing the concept that if both bounding functions converge to the same limit, so must f(x)f(x).

  5. Examples: Studying examples where the Squeeze Theorem applies can help solidify understanding, such as limits involving trigonometric functions or polynomial expressions.

  6. Conclusion: The Squeeze Theorem is a powerful tool in calculus for establishing limits, particularly when direct evaluation is challenging.

Focusing on these points will enhance your understanding of the Squeeze Theorem.

Part 2: Limit of sin(x)/x as x approaches 0

In this video, we prove that the limit of sin(θ)/θ as θ approaches 0 is equal to 1. We use a geometric construction involving a unit circle, triangles, and trigonometric functions. By comparing the areas of these triangles and applying the squeeze theorem, we demonstrate that the limit is indeed 1. This proof helps clarify a fundamental concept in calculus.

Here are the key points to learn when studying the limit of sin(x)x\frac{\sin(x)}{x} as xx approaches 0:

  1. Limit Definition: The limit is expressed as limx0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}.

  2. Value of the Limit: The result of this limit is 1.

  3. Graphical Interpretation: The graph of y=sin(x)y = \sin(x) and the line y=xy = x approaches each other as xx approaches 0.

  4. L'Hôpital's Rule: This limit can be evaluated using L'Hôpital's Rule, applicable when the limit is in the indeterminate form 00\frac{0}{0}. Take the derivative of the numerator and denominator.

  5. Squeeze Theorem: An alternative approach is to use the Squeeze Theorem. For small values of xx, sin(x)\sin(x) is squeezed between xcos(x)x \cos(x) and xx.

  6. Taylor Series Expansion: The Taylor series expansion of sin(x)\sin(x) around 0 can also help, showing that sin(x)xx36+O(x5)\sin(x) \approx x - \frac{x^3}{6} + O(x^5), leading to the limit.

  7. Applications: This limit is foundational in calculus and is frequently used in derivatives and integrals involving trigonometric functions.

Focusing on these points will help solidify your understanding of this important limit in calculus.

Part 3: Limit of (1-cos(x))/x as x approaches 0

In this video, we explore the limit of (1-cos(x))/x as x approaches 0 and show that it equals 0. We use the Pythagorean trigonometric identity, algebraic manipulation, and the known limit of sin(x)/x as x approaches 0 to prove this result. This concept is helpful for understanding the derivative of sin(x).

When studying the limit of 1cos(x)x\frac{1 - \cos(x)}{x} as xx approaches 0, here are the key points to learn:

  1. Identify the Form: The limit initially gives the form 00\frac{0}{0}, which is indeterminate.

  2. L'Hôpital's Rule: You can apply L'Hôpital's Rule, which states that if 00\frac{0}{0} or \frac{\infty}{\infty} occurs, you can differentiate the numerator and denominator:

    limx01cos(x)x=limx0sin(x)1=sin(0)=0.\lim_{x \to 0} \frac{1 - \cos(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{1} = \sin(0) = 0.
  3. Taylor Series Expansion: Another approach is to use the Taylor series expansion for cos(x)\cos(x):

    cos(x)1x22+O(x4).\cos(x) \approx 1 - \frac{x^2}{2} + O(x^4).

    Substituting this into the limit gives:

    1cos(x)x22.1 - \cos(x) \approx \frac{x^2}{2}.

    Thus, the limit simplifies to:

    limx0x22x=limx0x2=0.\lim_{x \to 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \to 0} \frac{x}{2} = 0.
  4. Understanding the Behavior: Recognize that as xx approaches 0, 1cos(x)1 - \cos(x) behaves like x22\frac{x^2}{2}, which helps in visualizing the limit.

  5. Geometric Interpretation: The limit can also be understood in the context of the unit circle, where 1cos(x)1 - \cos(x) represents the vertical distance in a right triangle formed by the angle xx.

By grasping these key points, you will have a comprehensive understanding of the limit 1cos(x)x\frac{1 - \cos(x)}{x} as xx approaches 0.