Problem 13 - Calculus for Biology and Medicine

Question

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin x \cos x}{x(1-x)} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Direct Substitution

Substitute the value of \( x = 0 \) directly into the limit expression: \[ \lim_{x \to 0} \frac{\sin x \cos x}{x(1-x)} = \frac{\sin 0 \cos 0}{0(1-0)} \]This gives \( \frac{0}{0} \), an indeterminate form, so direct substitution does not work.

Step 2: Rewrite the Expression

Let's rewrite the original expression as:\[ \frac{\sin x \cos x}{x(1-x)} = \frac{\sin x \cos x}{x} \cdot \frac{1}{1-x} \]This separates the equation into two parts: \( \frac{\sin x \cos x}{x} \) and \( \frac{1}{1-x} \).

Step 3: Apply the Limit to Each Part

Now, we apply the limit to each part separately:1. \( \lim_{x \to 0} \frac{\sin x \cos x}{x} \) 2. \( \lim_{x \to 0} \frac{1}{1-x} \)

Step 4: Solve \( \lim_{x \to 0} \frac{\sin x \cos x}{x} \)

We know the trigonometric identity: \( \sin x \cos x = \frac{1}{2}(\sin 2x) \). Therefore:\[ \frac{\sin x \cos x}{x} = \frac{1}{2} \cdot \frac{\sin 2x}{x} \]Use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). So,\[ \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \cdot 1 = 2 \]Thus,\[ \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin 2x}{x} = \frac{1}{2} \cdot 2 = 1 \]

Step 5: Solve \( \lim_{x \to 0} \frac{1}{1-x} \)

The limit of a simple rational function \( \frac{1}{1-x} \) as \( x \to 0 \) is straightforward:\[ \lim_{x \to 0} \frac{1}{1-x} = \frac{1}{1-0} = 1 \]

Step 6: Combine and Find the Final Answer

Now combine the results from Steps 4 and 5:\[ \lim_{x \to 0} \frac{\sin x \cos x}{x(1-x)} = 1 \times 1 = 1 \]

Key Concepts

Indeterminate Forms

When evaluating limits, we often encounter situations where direct substitution leads to expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. Indeterminate forms do not provide enough information to directly determine the limit because they suggest multiple possibilities. In our original problem, substituting \( x = 0 \) resulted in the expression \( \frac{0}{0} \), indicating an indeterminate form, which means that we cannot simply find the limit by substitution.

Whenever you face an indeterminate form, your goal should be to manipulate the expression in a way that eliminates the indeterminacy. This might involve algebraic manipulation, using trigonometric identities, or applying limit laws that make it possible to evaluate the expression effectively. In our exercise, we rewrote the expression to separate and evaluate each component individually, avoiding the direct indeterminacy.

Trigonometric Identities

Trigonometric identities are crucial tools for simplifying trigonometric expressions, especially in limits. For our problem, we utilized the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \). This transformation is a specific trigonometric identity that helps in rewriting the expression to make the limit easier to evaluate.

By expressing \( \sin x \cos x \) in terms of \( \sin 2x \), we reduce complexity and avoid the indeterminate form \( \frac{0}{0} \). Additionally, knowing the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) is essential in evaluating expressions involving sine functions. The identity allows you to transform products into more manageable forms, facilitating straightforward limit evaluation. Always keep a good grip on common trigonometric identities, as these can simplify not just limits but a wide range of problems.

Limit Laws

Limit laws are foundational principles that help calculate limits systematically. They provide rules on how to manage complex limits by breaking them into simpler parts. In the exercise, we applied limit laws to handle each segment of the rewritten expression separately.

Limit laws include:

• The Sum Law: \( \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \)
• The Product Law: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \)
• The Quotient Law: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \), assuming \( \lim_{x \to a} g(x) eq 0 \)

Applying these laws allowed us to move past the indeterminate form and evaluate \( \lim_{x \to 0} \frac{\sin x \cos x}{x} \) separately from \( \lim_{x \to 0} \frac{1}{1-x} \). By conducting separate evaluations and then multiplying the results, we found the final answer effortlessly. Mastering limit laws can make the challenging process of finding limits much smoother and more intuitive.