Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0}\left(\csc ^{2} x-\frac{1}{x^{2}}\right)^{2} $$

Short answer

This limit evaluates to 0 as \( x \rightarrow 0 \).

Step-by-step solution

Calculate the Indeterminate Form

To determine if we have an indeterminate form, first evaluate each function inside the limit as \( x \) approaches 0. Observe that as \( x \rightarrow 0 \), \( \csc(x) = \frac{1}{\sin(x)} \rightarrow \infty \), making \( \csc^2(x) - \frac{1}{x^2} \) an indeterminate form \( \frac{\infty}{\infty} \).

Apply l'Hôpital's Rule to Simplify

Since we have an indeterminate form, we can use l'Hôpital's Rule. Differentiate the numerator and denominator separately, knowing that \( \csc(x) = \frac{1}{\sin(x)} \), we differentiate it to find that the derivative of \( \csc^2(x) \) with respect to \( x \) is \(-2 \csc^2(x) \cot(x)\) and the derivative of \( \frac{1}{x^2} \) is \(-\frac{2}{x^3}\).

Reevaluate Limit After Differentiation

Substitute the differentiated expressions back into the limit problem: \(\lim _{x \to 0} \left(-2 \csc^2(x) \cot(x) + \frac{2}{x^3}\right).\) Rethink the limit as \( x \rightarrow 0 \). The terms inside the limit simplify further since \( \csc(x) = \frac{1}{\sin(x)} \) behaves like \( \frac{1}{x} \) and \( \cot(x) = \frac{\cos(x)}{\sin(x)} \), the product and the subtraction of similar terms still result in 0 after simplifications.

Evaluate the Final Limit

Using algebraic simplification and the understanding of trigonometric functions approaching small values, simplify the expression until no further indeterminate forms are present. Observe after simplifications that the expression approaches \( 0 \) finally.

Key concepts

Trigonometric Limits

In calculus, limits involving trigonometric functions are common and can often be challenging. Understanding these limits is essential to tackle various problems. Consider the expression \( \csc(x) = \frac{1}{\sin(x)} \). As \( x \) approaches 0, \( \sin(x) \) approaches 0 too, making \( \csc(x) \) head towards infinity. This is why limits involving \( \csc(x) \) when \( x \) approaches 0 can present unique challenges.

When approaching limits involving trigonometric functions like \( \csc^2(x) \), it's crucial to pair it with the behavior of other related functions like \( \sin(x) \) or \( \cos(x) \). These can help to simplify expressions and better understand how they behave near key values.

In the limit \( \lim _{x \rightarrow 0}\left(\csc ^{2} x-\frac{1}{x^{2}}\right)^{2} \), identifying the trigonometric nature of \( \csc(x) \) aids in questioning whether the limit is a true indeterminate form. Here, both \( \csc^2(x) \) and \( \frac{1}{x^2} \) become infinite, signifying a need for further evaluation to determine their interaction at the limit point.

Indeterminate Forms

Indeterminate forms are expressions that do not immediately indicate a particular value or a limit as a variable approaches a certain point. One of the most common indeterminate forms is \( \frac{\infty}{\infty} \), which emerges often in calculus when dealing with limits.

In the limit problem \( \lim _{x \rightarrow 0}\left(\csc ^{2} x-\frac{1}{x^{2}}\right)^{2} \), evaluating each component as \( x \) approaches 0 leads to an indeterminate form. This occurs because both \( \csc^2(x) \) and \( \frac{1}{x^2} \) behave similarly as \( x \to 0 \), each tending towards infinity.

Recognizing indeterminate forms is essential before applying techniques like l'Hôpital's Rule. They signal that further simplification or differentiation is necessary to resolve the ambiguity and determine the true behavior of the limit.

Calculus Problem Solving

Solving calculus problems often involves simplifying complex expressions and evaluating limits without directly substituting values. l'Hôpital's Rule is a valuable tool when dealing with indeterminate forms.

To apply l'Hôpital’s Rule, one must first confirm the presence of an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Only under these circumstances can the rule be applied. It involves differentiating the numerator and denominator separately to find the limit.

In the provided example, the limit \( \lim _{x \to 0} \left(\csc^2(x) \cot(x) + \frac{2}{x^3}\right) \) emerges after differentiation due to l'Hôpital's Rule. The focus then shifts to simplifying and properly evaluating the new expression. This helps in gradually understanding whether the limit approaches an existing number.

Calculus problem-solving requires a combination of algebraic manipulation, understanding function behavior, and strategic application of theorems and rules like l'Hôpital's, to handle complex limits efficiently.