/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Evaluate limit. $$\lim _{x \ri... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate limit. $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x}$$

Short Answer

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Question: Evaluate the limit of the function as x approaches 0: $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x}$$ Answer: The limit of the given function as x approaches 0 is \(-\frac{1}{2}\).

Step by step solution

01

Identify the indeterminate form when x approaches 0

By plugging x = 0 directly into the given function, we get: $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x} = \frac{\cos (0)-1}{\sin^2(0)} = \frac{1-1}{0}=\frac{0}{0}$$ Since we have an indeterminate form (0/0), we cannot directly evaluate the limit. Therefore, we need to use L'Hôpital's Rule.
02

Apply L'Hôpital's Rule

Before applying L'Hôpital's Rule, let's ensure that the given conditions are satisfied. L'Hôpital's Rule can be applied if the limits of the derivatives of the numerator and the denominator exist. So, let's differentiate both the numerator and the denominator. Differentiating the numerator \(\cos x - 1\) with respect to x, we get: $$\frac{d}{dx}(\cos x-1)=-\sin x$$ Differentiating the denominator \(\sin^2x\) with respect to x, we get: $$\frac{d}{dx}(\sin ^{2} x)=2\sin x\cos x$$ Now, we can apply L'Hôpital's Rule: $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{-\sin x}{2\sin x\cos x}$$
03

Simplify the expression and evaluate the limit

Now, let's simplify the expression and evaluate the limit again. Divide both numerator and denominator by \(\sin x\) (keeping in mind that as x approaches 0, sin(x) is not equal to 0): $$\lim _{x \rightarrow 0} \frac{-\sin x}{2\sin x\cos x}=\lim _{x \rightarrow 0} \frac{-1}{2\cos x}$$ Now, evaluate the limit: $$\lim _{x \rightarrow 0} \frac{-1}{2\cos x} = \frac{-1}{2\cos (0)} = \frac{-1}{2(1)} = -\frac{1}{2}$$ The limit of the given function as x approaches 0 is \(-\frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hôpital's Rule
L'Hôpital's Rule is an essential tool in calculus used to resolve limits that present an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter these forms, direct substitution into the limit does not work because they do not yield a definitive result. Furthermore, indeterminate forms provide an indication that further analysis or manipulation of the function is necessary. One effective method is to use L'Hôpital's Rule, which can simplify evaluating these troublesome limits.

Here's how it works:
  • First, ensure that the limit initially results in either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
  • Next, take the derivative of both the numerator and the denominator of the function separately.
  • After differentiating, evaluate the limit again.
  • If necessary, repeat the process until the limit no longer results in an indeterminate form.
In the original exercise, following these steps helped resolve the issue, allowing the limit to be simplified and evaluated properly.
Indeterminate Forms
An indeterminate form in mathematics is an expression that doesn't lead directly to a result when substituting a specific point into a function. Common examples include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and more. They are labeled as 'indeterminate' because we cannot ascertain their values through direct substitution.

Recognizing an indeterminate form is crucial when evaluating limits, as it indicates that you cannot rely on straightforward computation. Instead, alternative techniques, such as algebraic manipulation or applying rules like L'Hôpital's Rule, are required. In situations where you find indeterminate forms:
  • Pause and verify the presence of an indeterminate condition by substituting the limit point into the function.
  • Determine the best approach (e.g., factorization, substitution, or using L'Hôpital's Rule) to resolve the limit.
  • Test the new function or form by substituting values to ensure that the indeterminate form is resolved.
In the exercise discussed, \(\frac{0}{0}\) represents an indeterminate form which tells us that a deeper analysis is needed for accurate evaluation.
Trigonometric Limits
Trigonometric limits often deal with functions involving sine, cosine, and other trigonometric relationships, which are essential in calculus and real-world applications. Understanding trigonometric limits involves recognizing certain well-known properties and limits.

A fundamental limit to recall is \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). This and similar standard limits are helpful as building blocks for more complex evaluations involving trigonometric functions.
  • Learn and memorize basic trigonometric limits, as they frequently appear in calculus problems.
  • Utilize trigonometric identities to simplify expressions, making limits easier to evaluate.
  • Consider rewriting complex expressions in simpler trigonometric terms when possible.
In our exercise, the presence of \(\sin^2(x)\) and \(\cos(x)-1\) stresses the importance of a firm grasp on trigonometric identities and simplifications. Evaluating the limit involved understanding these trigonometric properties alongside calculus techniques like L'Hôpital's Rule.

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