Limits and Continuity

Limits Involving Trigonometric Functions

Evaluating limits that involve trigonometric functions often requires the use of trigonometric identities and standard limit results. One common technique is to rewrite the expression using identities to simplify the limit calculation.

• Key Identity: The double-angle identity for cosine: $\cos x = 1 - 2 \sin^2 \frac{x}{2}$

• Standard Limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Example: Evaluate $\lim_{x \to 0} \frac{-2 \sin^2 \frac{x}{2}}{x}$

• Rewrite the numerator using the double-angle identity:

• $\cos x = 1 - 2 \sin^2 \frac{x}{2} \implies -2 \sin^2 \frac{x}{2} = \cos x - 1$

• So, $\lim_{x \to 0} \frac{-2 \sin^2 \frac{x}{2}}{x}$

• Break up the limit:

• $= \lim_{x \to 0} -2 \cdot \frac{\sin^2 \frac{x}{2}}{x}$

• Express $\sin^2 \frac{x}{2}$ as $\left(\sin \frac{x}{2}\right)^2$:

• $= \lim_{x \to 0} -2 \cdot \frac{\sin \frac{x}{2}}{x} \cdot \sin \frac{x}{2}$

• Rewrite $\frac{\sin \frac{x}{2}}{x}$ as $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \cdot \frac{1}{2}$:

• $= \lim_{x \to 0} -2 \cdot \left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \frac{1}{4}$

• As $x \to 0$, $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \to 1$:

• $= -2 \cdot 1^2 \cdot \frac{1}{4} = -\frac{1}{2}$

Conclusion: The limit evaluates to $-\frac{1}{2}$.

Additional info: This example demonstrates the use of trigonometric identities and the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to evaluate more complex limits involving trigonometric functions.