Problem 9 - Calculus for Biology and Medicine

Question

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin (\pi x)}{x} $$

Step-by-Step Solution

Answer

The limit evaluates to \( \pi \).

Step 1: Recognize the Standard Limit Form

We need to evaluate \[ \lim _{x \rightarrow 0} \frac{\sin (\pi x)}{x}. \]This resembles the standard limit form for trigonometric functions: \[ \lim _{x \rightarrow 0} \frac{\sin(x)}{x} = 1. \]Our goal is to use this standard form by adjusting our given limit to fit its structure.

Step 2: Substitute and Transform the Expression

To utilize the standard limit form, set \( u = \pi x \). Then as \( x \rightarrow 0 \), \( u \rightarrow 0 \) as well. Thus, the expression becomes:\[ \lim _{u \rightarrow 0} \frac{\sin(u)}{u/\pi} = \pi \cdot \lim _{u \rightarrow 0} \frac{\sin(u)}{u}. \]

Step 3: Apply the Known Limit

We know from the standard trigonometric limit that\[ \lim _{u \rightarrow 0} \frac{\sin(u)}{u} = 1. \]Thus, substituting this result into our expression gives:\[ \pi \cdot 1 = \pi. \]

Key Concepts

Standard Limit Form

When dealing with limits in trigonometry, especially those involving sine and tangent, the **standard limit form** is a powerful tool. Recognizing these forms helps in simplifying and solving problems quickly. The most common of these is:

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

When your limit problem resembles this structure, it can often be rewritten or transformed to directly use this standard form. By utilizing this formula, you can avoid complex computations and derive the answer in just a couple of steps. It's important to employ this standard form whenever you encounter limits of sine and similar trigonometric functions approaching zero, making it an essential part of limit evaluation.

Limit Evaluation

**Limit evaluation** in trigonometry involves calculating the value that a function approaches as the input approaches a certain point. Here's how you can effectively evaluate a trigonometric limit:

• Firstly, compare the given limit with known standard forms. If it matches or can be matched, you have made a significant step towards solving it.
• Next, properly set up the limit to see if direct substitution or some manipulation is needed. This ensures the problem is suited for the application of known limits or other methods.

In the exercise, the limit \( \lim_{x \rightarrow 0} \frac{\sin(\pi x)}{x} \) was recognized as a candidate for the standard form. By identifying structural similarities with \( \lim_{x \to 0} \frac{\sin(x)}{x} \), the problem becomes more approachable, streamlining the limit evaluation process.

Substitution in Limits

One of the fundamental techniques in solving trigonometric limits is through **substitution in limits**. This involves altering the variable or the form of the limit to make it easier to apply known formulas or standard forms. In this exercise, the substitution was:

• Set \( u = \pi x \), then as \( x \to 0 \), \( u \to 0 \).

This transforms the original expression into something potentially easier to handle:

• \( \lim_{u \rightarrow 0} \frac{\sin(u)}{u/\pi} \)

By expressing \( u \) in terms of \( x \), we can exploit the standard limit form. Thus, \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \) was applied directly, ultimately yielding the final result by multiplying by constant factors introduced by the substitution. Substitution is an invaluable approach for aligning more complex expressions with simpler, known ones.