Problem 29 - Thomas Calculus

Question

Find the limits in Exercises \(21-42\) $$\lim _{x \rightarrow 0} \frac{x+x \cos x}{\sin x \cos x}$$

Step-by-Step Solution

Answer

The limit does not exist because \( \frac{1}{\sin x} \) approaches infinity as \( x \to 0 \).

Step 1: Identify the expression

The given limit expression is \( \lim _{x \rightarrow 0} \frac{x + x \cos x}{\sin x \cos x} \).

Step 2: Simplify the numerator

Factor \( x \) out of the numerator: \( x + x \cos x = x(1 + \cos x) \). Now the expression becomes \( \frac{x(1 + \cos x)}{\sin x \cos x} \).

Step 3: Simplify the expression

The expression simplifies to \( \frac{1+\cos x}{\sin x} \) because the \( x \) in the numerator and denominator cancels out.

Step 4: Apply limit properties

Find \( \lim_{x \to 0} \frac{1 + \cos x}{\sin x} \). This can be rewritten using limits: \( \lim_{x \to 0} (1 + \cos x) \times \lim_{x \to 0} \frac{1}{\sin x} \).

Step 5: Evaluate the limits separately

Evaluate \( \lim_{x \to 0} (1 + \cos x) = 1 + \cos(0) = 2 \). Evaluate \( \lim_{x \to 0} \frac{1}{\sin x} \).Notice that \( \frac{1}{\sin x} \) approaches infinity as \( \sin x \to 0 \). Thus, the expression doesn't have a finite limit.

Key Concepts

Trigonometric Limits

Trigonometric limits are key in understanding the behavior of trigonometric functions as they approach specific points. These limits often involve functions like sine, cosine, and tangent.
In the problem above, the expression involves sine and cosine functions. Studying these helps determine how the expression behaves as it nears zero.

• Sine \(\sin x\): As \(x \to 0\), \(\sin x\) approaches \(0\). This is crucial since the denominator involves \(\sin x\).
• Cosine \(\cos x\): At zero, \(\cos x\) equals \(1\). This affects the numerator as \(x \times (1 + \cos x)\) simplifies the problem.

Understanding these trigonometric functions is fundamental to simplifying the expression and finding the limit.

Limit Evaluation Techniques

Knowing various limit evaluation techniques allows for the simplification and evaluation of expressions that initially pose challenges. Here, several methods are used:

• Factoring: The first step is factoring \(x\) out of the numerator. This simplifies the expression and allows for the cancellation of terms.
• Cancellation: Cancelling \(x\) in the numerator and denominator simplifies \(\frac{x(1 + \cos x)}{\sin x \cos x}\) to \(\frac{1 + \cos x}{\sin x}\).
• Direct Substitution: Though initially challenging due to \(\frac{0}{0}\), simplifying results in evaluating limits individually, like \((1 + \cos x)\) and \(\frac{1}{\sin x}\).

These techniques are vital for evaluating complex expressions and making expressions more manageable.

Indeterminate Forms

Indeterminate forms often arise when evaluating limits, causing the expression to appear undefined initially. Common forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and \(0\cdot\infty\).
In the given problem, the limit initially presents a \(\frac{0}{0}\) form. This signals further simplification is necessary.

• Simplification: Simplifying the expression changes its form from indeterminate to one that can be evaluated through limit laws.
• Breaking Down the Limit: After cancelling \(x\), the limit becomes \(\frac{1 + \cos x}{\sin x}\), heading towards a solution.

Addressing indeterminate forms requires skill and understanding of calculus concepts to resolve and interpret limits correctly.