Squeeze theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is a crucial concept in calculus, especially when dealing with limits. Here are the key points to learn when studying the Squeeze Theorem:

1. Definition: The theorem states that if you have three functions $f(x)$, $g(x)$, and $h(x)$ such that $f(x) \leq g(x) \leq h(x)$ for all $x$ in some interval around a point (except possibly at the point itself), and if $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} h(x) = L$, then $\lim_{x \to c} g(x) = L$.

2. Application: The Squeeze Theorem is useful for finding limits of functions that are difficult to evaluate directly. It helps to bound the function of interest between two other functions where limits are easier to compute.

3. Conditions:

   • The functions must be comparable (i.e., $f(x) \leq g(x) \leq h(x)$).
   • The limits of the bounding functions must exist and be equal as $x$ approaches a particular value.

4. Examples: Common examples include limits involving trigonometric functions, such as proving that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ by squeezing it between $\cos(x)$ and $1$.

5. Graphical Interpretation: Visualizing the functions can help understand how squeezing works. The graph of $g(x)$ must be trapped between the graphs of $f(x)$ and $h(x)$.

6. Limit Notation: Be comfortable using limit notation and the symbols indicating that $x$ approaches a value, as this is integral to applying the theorem correctly.

By mastering these points, you will be well-prepared to use the Squeeze Theorem effectively in limit problems.

The key points to understand when studying the limit of $\frac{\sin(x)}{x}$ as $x$ approaches 0 are:

1. Limit Value: The limit is $1$. That is, $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$.

2. Squeeze Theorem: The limit can be demonstrated using the Squeeze Theorem by comparing $\sin(x)$ with $x$ and $\sin(x) \leq x \leq \tan(x)$ for small values of $x$.

3. Taylor Series Expansion: The Taylor series expansion of $\sin(x)$ around $x = 0$ shows that $\sin(x) \approx x$ as $x$ approaches 0.

4. Geometric Interpretation: A geometric approach involves considering the unit circle to visualize the relationship between $\sin(x)$ and the arc length.

5. L'Hôpital's Rule: As $\frac{0}{0}$ is an indeterminate form, applying L'Hôpital's Rule (differentiating numerator and denominator) also yields the limit as $1$.

6. Continuity of Functions: Understanding that $\sin(x)$ is continuous and differentiable helps in comprehending the behavior of the function around $x = 0$.

These points collectively reinforce the understanding of this fundamental limit in calculus.

When studying the limit of $(1 - \cos(x))/x$ as $x$ approaches 0, here are the key points to consider:

1. Understanding the Limit: The expression tends toward an indeterminate form $\frac{0}{0}$ as $x$ approaches 0, necessitating further analysis.

2. Taylor Series Expansion: Use the Taylor series expansion of $\cos(x)$ around $x = 0$:

   $$\cos(x) = 1 - \frac{x^2}{2} + O(x^4)$$

   This allows you to rewrite $1 - \cos(x)$ as:

   $$1 - \cos(x) \approx \frac{x^2}{2} \quad \text{for small } x$$

3. Substituting Back into the Limit: Substitute this approximation back into the limit:

   $$\lim_{x \to 0} \frac{1 - \cos(x)}{x} \approx \lim_{x \to 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \to 0} \frac{x}{2} = 0$$

4. L'Hôpital's Rule: Alternatively, apply L'Hôpital's Rule since the limit is in $\frac{0}{0}$ form. Differentiate the numerator and the denominator:

   $$\text{Numerator: } \frac{d}{dx}(1 - \cos(x)) = \sin(x)$$
   $$\text{Denominator: } \frac{d}{dx}(x) = 1$$

   This gives:

   $$\lim_{x \to 0} \frac{\sin(x)}{1} = \sin(0) = 0$$

5. Conclusion: From both methods, you conclude that:

   $$\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$$

By exploring these points, you develop a deeper understanding of limits involving trigonometric functions and can apply similar techniques to other limits as well.