Limit of Trigonometric Function

Understanding Limits of Trigonometric Functions

Imagine you observe a clock pendulum swinging very slowly approaching its equilibrium point. This motion is similar to the behavior of trigonometric functions when their variable approaches a certain value. Limits of trigonometric functions examine how the values of sine, cosine, and tangent functions behave when the input approaches critical points.

Unlike limits of algebraic functions that can often be solved by direct substitution, trigonometric functions have special characteristics due to their periodic and oscillating nature. This makes us need to use theorems and special properties to solve trigonometric limits.

Fundamental Theorem of Trigonometric Limits

The most important foundation in trigonometric limits is the theorem stating that the sine function approaches its gradient when the angle approaches zero.

Basic Sine Limit

The most fundamental theorem is:

This theorem cannot be proven using direct substitution because it results in the form $\frac{0}{0}$. Its proof requires a geometric approach using the unit circle and the squeeze theorem.

In this theorem, $x$ must be in radians, not degrees. If using degrees, the result will be different.

Consequences of the Basic Limit

From the fundamental limit above, we can derive several other important limits:

Properties of Trigonometric Limits

Based on the fundamental theorem, we can build a series of very useful properties:

Trigonometric Ratios

For constants $a \neq 0$ and $b \neq 0$, and all following limits when $x \to 0$:

Trigonometric Combinations

From the trigonometric ratio properties, we can derive several trigonometric combination properties:

Techniques for Solving Trigonometric Limits

Substitution and Manipulation Techniques

When facing complex trigonometric limits, we often need to manipulate expressions to use the fundamental theorem.

Calculate: $\lim_{x \to 0} \frac{\sin 2x}{3x}$

We manipulate to obtain the standard form:

Trigonometric Identity Techniques

Often we need to use trigonometric identities to simplify expressions. Always ensure the function is defined at the point being approached.

Calculate: $\lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x}$

Since $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \neq 0$, the denominator is not zero at $x = \frac{\pi}{4}$.

Direct substitution:

Techniques for Special Forms

For limits involving the form $\frac{0}{0}$, we need special techniques.

Calculate: $\lim_{x \to \frac{\pi}{2}} \frac{\sin^2(x - \frac{\pi}{2})}{x(x - \frac{\pi}{2})}$

Let $u = x - \frac{\pi}{2}$, then when $x \to \frac{\pi}{2}$, we have $u \to 0$ and $x = u + \frac{\pi}{2}$.

Trigonometric Limits with Angle Identities

Using Sum and Difference Formulas

When dealing with trigonometric functions involving sum or difference of angles, we can use trigonometric identities.

Calculate: $\lim_{t \to 0} \frac{\cot 5t}{\cot 10t}$

Using the definition of cotangent:

Exercises

1. Calculate $\lim_{x \to 0} \frac{\sin 3x}{4x}$

2. Calculate $\lim_{x \to 0} \frac{\tan 2x}{\sin 5x}$

3. Calculate $\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}$

4. Calculate $\lim_{x \to \frac{\pi}{6}} \frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}$

5. Calculate $\lim_{x \to 0} \frac{\sin x \cos x}{x}$

Answer Key

1. Solution:

   Use algebraic manipulation to obtain the standard form:

   $$\lim_{x \to 0} \frac{\sin 3x}{4x} = \lim_{x \to 0} \frac{3 \sin 3x}{3 \cdot 4x} = \lim_{x \to 0} \frac{3}{4} \cdot \frac{\sin 3x}{3x}$$

   $$= \frac{3}{4} \lim_{x \to 0} \frac{\sin 3x}{3x} = \frac{3}{4} \cdot 1 = \frac{3}{4}$$

2. Solution:

   Use trigonometric ratio properties:

   $$\lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \frac{5x}{\sin 5x} \cdot \frac{2x}{5x}$$

   $$= \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \lim_{x \to 0} \frac{5x}{\sin 5x} \cdot \frac{2}{5}$$

   $$= 1 \cdot 1 \cdot \frac{2}{5} = \frac{2}{5}$$

3. Solution:

   Use the identity $1 - \cos ax = 2\sin^2\frac{ax}{2}$:

   $$\lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{2\sin^2\frac{3x}{2}}{x^2}$$

   $$= \lim_{x \to 0} 2 \cdot \left(\frac{\sin\frac{3x}{2}}{\frac{3x}{2}}\right)^2 \cdot \left(\frac{3}{2}\right)^2$$

   $$= 2 \cdot 1^2 \cdot \frac{9}{4} = \frac{9}{2}$$

4. Solution:

   Let $h = x - \frac{\pi}{6}$, then $x = h + \frac{\pi}{6}$ and when $x \to \frac{\pi}{6}$, we have $h \to 0$:

   $$\lim_{h \to 0} \frac{\sin(h + \frac{\pi}{6}) - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\sin h \cos\frac{\pi}{6} + \cos h \sin\frac{\pi}{6} - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\sin h \cdot \frac{\sqrt{3}}{2} + \cos h \cdot \frac{1}{2} - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\frac{\sqrt{3}}{2}\sin h + \frac{1}{2}(\cos h - 1)}{h}$$

   $$= \frac{\sqrt{3}}{2} \lim_{h \to 0} \frac{\sin h}{h} + \frac{1}{2} \lim_{h \to 0} \frac{\cos h - 1}{h}$$

   $$= \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2} \cdot 0 = \frac{\sqrt{3}}{2}$$

5. Solution:

   Use the identity $\sin x \cos x = \frac{1}{2}\sin 2x$:

   $$\lim_{x \to 0} \frac{\sin x \cos x}{x} = \lim_{x \to 0} \frac{\frac{1}{2}\sin 2x}{x}$$

   $$= \frac{1}{2} \lim_{x \to 0} \frac{\sin 2x}{x} = \frac{1}{2} \lim_{x \to 0} \frac{2 \sin 2x}{2x}$$

   $$= \frac{1}{2} \cdot 2 \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \cdot 1 = 1$$