Squeeze Theorem and Special Limits

When evaluating limits in calculus, you will sometimes encounter functions that oscillate wildly or cannot be simplified using standard algebraic methods. In these cases, the Squeeze Theorem (also known as the Sandwich Theorem) and Special Trigonometric Limits become essential tools.

The Squeeze Theorem

The Squeeze Theorem states that if you can trap a complex function between two simpler functions that both approach the same limit, the complex function must also approach that same limit.

Mathematically, if $f(x) \le g(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$), and: $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$ Then it must be true that: $\lim_{x \to a} g(x) = L$

Example: Using the Squeeze Theorem

Find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

1. Bound the oscillating part: We know that the sine function, no matter its argument, always outputs values between $-1$ and $1$. $-1 \le \sin\left(\frac{1}{x}\right) \le 1$
2. Construct the original function: Multiply all parts of the inequality by $x^2$. Since $x^2 \ge 0$, the inequality signs do not flip. $-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$
3. Evaluate the outer limits: Take the limit as $x \to 0$ for the outer functions $f(x) = -x^2$ and $h(x) = x^2$. $\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0$
4. Conclusion: Because the outer functions both approach $0$, the inner function is "squeezed" to $0$ as well. $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

Special Trigonometric Limits

The Squeeze Theorem is used to prove one of the most important foundational limits in calculus: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

From this, another crucial special limit is derived: $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

These special limits allow us to evaluate more complex trigonometric expressions by algebraically manipulating them into recognizable forms.

Example: Using Special Limits

Evaluate $\lim_{x \to 0} \frac{\tan(3x)}{2x}$

1. Rewrite tangent: Express $\tan(3x)$ as $\frac{\sin(3x)}{\cos(3x)}$. $\lim_{x \to 0} \frac{\sin(3x)}{2x \cos(3x)}$
2. Rearrange the expression: We want to create the form $\frac{\sin u}{u}$, where $u = 3x$. $\lim_{x \to 0} \left( \frac{\sin(3x)}{x} \cdot \frac{1}{2 \cos(3x)} \right)$
3. Manipulate to match the special limit: Multiply the numerator and denominator by $3$ to match the argument of the sine function. $\lim_{x \to 0} \left( \frac{3 \sin(3x)}{3x} \cdot \frac{1}{2 \cos(3x)} \right)$
4. Evaluate the limit: Apply the special limit $\lim_{3x \to 0} \frac{\sin(3x)}{3x} = 1$ and substitute $x = 0$ into the remaining cosine term (since $\cos(0) = 1$). $3(1) \cdot \frac{1}{2(1)} = \frac{3}{2}$