Problem 33 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0}(\csc x-1 / x)$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Rewrite the Expression

To find the limit of \(\csc x - \frac{1}{x}\) as \(x \to 0\), start by expressing \(\csc x\) in terms of sine: \(\csc x = \frac{1}{\sin x}\). Thus, the expression becomes \(\frac{1}{\sin x} - \frac{1}{x}\).

Step 2: Combine into a Single Fraction

Combine the expression \(\frac{1}{\sin x} - \frac{1}{x}\) into a single fraction: \(\frac{x - \sin x}{x \sin x}\).

Step 3: Apply L'Hôpital's Rule

Evaluate the limit \(\lim_{x \rightarrow 0} \frac{x - \sin x}{x \sin x}\). Since this is an indeterminate form \(\frac{0}{0}\), apply L'Hôpital's Rule, which states to take the derivatives of the numerator and the denominator.

Step 4: Derivatives of Numerator and Denominator

The derivative of the numerator \(x - \sin x\) is \(1 - \cos x\). The derivative of the denominator \(x \sin x\) is \(\sin x + x \cos x\).

Step 5: Apply L'Hôpital's Rule Second Time

Compute the limit using the derivatives: \(\lim_{x \rightarrow 0} \frac{1 - \cos x}{\sin x + x \cos x}\). Substitute \(x = 0\) to get \(\frac{1 - 1}{0 + 0} = \frac{0}{0}\), which is still indeterminate, so apply L'Hôpital's Rule again.

Step 6: Second Derivatives

Find second derivatives. For the numerator \(1 - \cos x\), the derivative is \(\sin x\). For the denominator \(\sin x + x \cos x\), the derivative is \(\cos x - \sin x - x \sin x\).

Step 7: Final Limit Calculation

Evaluate the limit \(\lim_{x \rightarrow 0} \frac{\sin x}{\cos x + x \cos x - \sin x}\). At \(x = 0\), this yields \(\frac{0}{1} = 0\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool in calculus, particularly when dealing with limits involving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When a limit problem presents an indeterminate form, this rule helps us by simplifying the evaluation process.

The rule states that if the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can find the limit instead by taking the derivatives of the numerator \( f(x) \) and the denominator \( g(x) \), and then evaluating \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \). This can be repeated if the new limit also results in an indeterminate form.

• Ensure that both numerator and denominator are differentiable in the neighborhood of \( a \).
• Differentiate the numerator and the denominator separately.
• Apply the rule repeatedly as needed until the indeterminate form is resolved.

Although powerful, L'Hôpital's Rule can only be applied under the conditions specified above.

Trigonometric Limits

Trigonometric limits often involve functions like sine, cosine, and other related trigonometric expressions. Understanding these limits is key in calculus, particularly because trigonometric functions have unique properties around certain values like 0 or \(\pi\).

One classic limit often encountered is \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which arises when examining the behavior of sine near zero. This is particularly applicable in simplifying expressions where trigonometric functions might otherwise complicate the evaluation of limits.

• Simplify complex trigonometric expressions into simpler forms if possible.
• Utilize known trigonometric limits to aid in computations.
• Examine the special values of \( \pi \), \( \frac{\pi}{2} \), and their multiples, where trigonometric functions often change behavior significantly.

When combined with L'Hôpital's Rule, trigonometric limits can often be resolved elegantly, as is seen in solving exercises involving limits of trigonometric differences.

Indeterminate Forms

Indeterminate forms occur in calculus when direct substitution in a limit problem doesn't yield a clear result, often presenting as expressions like \( \frac{0}{0} \) or \( \infty - \infty \). These forms require additional manipulation or tools like L'Hôpital's Rule to find the actual limit.

When faced with an indeterminate form, the goal is to transform the expression into something more manageable that can be effectively evaluated. Here's a common strategy:

• Reformulate the expression to reveal any hidden cancellations or transformations that simplify it.
• Consider multiplying or dividing by conjugates or factoring to simplify complex expressions.
• Apply algebraic techniques like partial fractions or polynomial division as necessary.

Identifying and resolving indeterminate forms is crucial for correctly evaluating limits, and they often act as a signal to employ calculus techniques like differentiation or series expansion.