Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow \pi} \frac{\sin n x}{\sin x} $$

Short answer

Function: \(\frac{\sin nx}{\sin x}\) Answer: The limit of the function as x approaches π is \(\cos( (n-1)\pi)\).

Step-by-step solution

Identify the trigonometric identity

We will use the sine angle addition formula: \(\sin(a + b) = \sin a \cos b + \cos a \sin b\)

Apply the sine angle addition formula

Using the sine angle addition formula, we can rewrite the function in a more manageable form. Let's represent \(a = (n - 1)x\) and \(b = x\): \[ \frac{\sin nx}{\sin x} = \frac{\sin((n-1)x + x)}{\sin x} = \frac{\sin( (n-1)x)\cos x + \cos( (n-1)x)\sin x}{\sin x} \]

Split the fraction

Now, we can split the fraction into two parts: \[ \frac{\sin nx}{\sin x} = \frac{\sin( (n-1)x)\cos x}{\sin x} + \frac{\cos( (n-1)x)\sin x}{\sin x} \]

Simplify and find the limit

The second term becomes simply the cos((n-1)x) since the sin x in the numerator and denominator will cancel out. Now the expression becomes: \[ \frac{\sin nx}{\sin x} = \frac{\sin( (n-1)x)\cos x}{\sin x} + \cos( (n-1)x) \] Now, find the limit as \(x \rightarrow \pi\): \[ \lim_{x \rightarrow \pi} \left(\frac{\sin( (n-1)x)\cos x}{\sin x} + \cos( (n-1)x)\right) \] Since the functions are continuous at \(x = \pi\), we can substitute the value of \(\pi\) into the expression: \[ \frac{\sin( (n-1)\pi)\cos \pi}{\sin \pi} + \cos( (n-1)\pi) \]

Evaluate the trigonometric functions and simplify

Evaluate the sine and cosine functions: \[ \frac{\sin( (n-1)\pi)(-1)}{0} + \cos( (n-1)\pi) \] We can see that for the first term, the denominator is 0, so it becomes undefined. For the second term, cosine function at \(((n-1)\pi)\) oscillates between 1 and -1 for integer values of n. Therefore, the limit of the function as \(x \rightarrow \pi\) is: \[ \lim_{x \rightarrow \pi} \frac{\sin nx}{\sin x} = \cos( (n-1)\pi) \]

Key concepts

Trigonometric Limits

When working with limits in calculus, especially trigonometric limits, understanding how specific functions behave as they approach a particular point is essential. Limits involving trigonometric functions often require special techniques or identities to solve.
In our case, we had to evaluate \( \lim_{x \rightarrow \pi} \frac{\sin nx}{\sin x} \).

• As \(x\) approaches \(\pi\), the trigonometric nature of \(\sin\) and \(\cos\) creates complexities that often result in indeterminate forms like \(\frac{0}{0}\).
• This requires rewriting or manipulating the function to reveal clearer behavior. Our method involved using a trigonometric identity to simplify the expression.
• The sine angle addition formula helped split and reorganize the expression, setting up the problem for a successful limit evaluation.

The key to mastering trigonometric limits is practice and familiarity with trig identities, which lay the groundwork for unraveling complex expressions.

Sine and Cosine Functions

Sine and cosine functions are fundamental in trigonometry and calculus. These periodic functions have properties that are extensively used in solving limits.
The identity \( \sin(a + b) = \sin a \cos b + \cos a \sin b \) is just one example of how sine and cosine can be used to break down more complex expressions.

• Sine and cosine functions oscillate between -1 and 1, and are continuous everywhere, which means they don't have gaps or jumps.
• Their periodic nature can be advantageous: e.g., for \(\cos((n-1)\pi)\), the cosine function naturally helps simplify the expression since its output oscillates predictably based on the value of \(n\).
• Such properties are crucial when we substitute \(x\) with specific values like \(\pi\) to determine limits.

Mastery of these functions is vital for calculus as they reveal patterns and shortcuts when evaluating trigonometric limits.

Continuous Functions

Continuous functions are functions that are smooth and connected; they do not have breaks or jumps. In calculus, understanding continuity aids in calculating limits, as it allows you to directly substitute values when finding limits, assuming no division by zero occurs.
In our scenario, evaluating \(\lim_{x \rightarrow \pi} \frac{\sin nx}{\sin x} \) involved continuous sine and cosine functions.

• The sine and cosine functions are beautifully continuous, providing reliable behavior at \(x = \pi\).
• Given continuity, once the expression is simplified, we can safely substitute the limiting value \(x = \pi\) into the expression directly.
• However, careful measures must be taken to avoid situations where division by zero could falsely seem acceptable due to continuity alone.

Comprehending continuous functions helps streamline limit evaluations by ensuring valid substitution is utilized when simplifying complex expressions.