/*! 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 48 Find the limits. $$\lim _{x \r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the limits. $$\lim _{x \rightarrow 0}\left(x^{2}-1\right)(2-\cos x)$$

Short Answer

Expert verified
The limit is -1.

Step by step solution

01

Understand the problem

We need to find the limit of the expression \((x^2 - 1)(2 - \cos x)\) as \(x\) approaches 0. This requires substitution and simplification.
02

Substitute the limit value

Substitute \(x = 0\) into the expression, we get: \((0^2 - 1)(2 - \cos(0)) = (-1)(2 - 1) = -1.\) We directly substitute because the expression is continuous at \(x = 0\).
03

Verify continuity

The expression \((x^2 - 1)(2 - \cos x)\) is a product of continuous functions. Since both \(x^2 - 1\) and \(2 - \cos x\) are continuous at \(x = 0\), the entire expression is continuous at \(x = 0\). Thus, the limit is simply the value of the function at that point.

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Key Concepts

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

Continuity in Mathematics
Continuity is a fundamental concept in calculus and analysis, which describes a condition where small changes in the input of a function lead to small changes in the output. For a function to be continuous at a point, specifically at \(x = c\), three conditions must be satisfied:
  • The function \(f(x)\) must be defined at \(x = c\).
  • The limit \(\lim_{x \to c} f(x)\) must exist.
  • The limit must be equal to the function's value: \(f(c) = \lim_{x \to c} f(x)\).
When these conditions are met, the function can be graphed without any breaks or gaps at that point. Continuous functions exhibit predictable behaviors which simplifies the process of analyzing them mathematically. In this exercise, the expression \((x^2 - 1)(2 - \cos x)\) was continuous at \(x = 0\), allowing us to evaluate the limit via direct substitution without complexity. This is because both components, \(x^2 - 1\) and \(2 - \cos x\), are continuous functions and retain their continuity when multiplied together.
Substitution Method in Limits
The substitution method allows us to find limits by directly replacing the variable with its limiting value. This method can be used when a function is continuous at the point we're evaluating. In our exercise, to find \( \lim_{x \to 0} (x^2 - 1)(2 - \cos x)\), we substituted \(x = 0\) into the expression:
  • The term \(x^2 - 1\) becomes \(0^2 - 1 = -1\).
  • The term \(2 - \cos x\) becomes \(2 - \cos(0) = 2 - 1 = 1\).
Combining these results gives \((-1)(1) = -1\). The substitution method works seamlessly when the conditions of continuity are satisfied, making it a straightforward approach to solve problems involving limits.
Limit Evaluation Techniques
Evaluating limits often involves various techniques that ensure we accurately measure a function's behavior as it approaches a specific point. Some key techniques include:
  • **Direct Substitution:** Used primarily for continuous functions, where the variable is directly replaced with the limiting value, like in our example with \(x = 0\).
  • **Factoring and Simplification:** These techniques handle more complex expressions by reducing them to simpler forms before evaluating.
  • **Rationalization:** Commonly used for expressions involving roots, this technique involves multiplying by the conjugate to simplify.
  • **L'Hôpital's Rule:** This powerful tool applies to indeterminate forms like \(0/0\) or \(\infty/\infty\), using derivatives to find the limit.
Our specific exercise did not require advanced techniques beyond direct substitution due to the continuity of the function. However, understanding these methods is fundamental for tackling a wide range of limit problems in calculus, providing students with a comprehensive toolbox for problem-solving.

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Most popular questions from this chapter

Give an example of a function \(f(x)\) that is continuous for all values of \(x\) except \(x=2\). where it has a removable discontinuity. Explain how you know that \(f\) is discontinuous at \(x=2,\) and how you know the discontinuity is removable.

Explain why the equation \(\cos x=x\) has at least one solution.

Graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$f(x)=\frac{\sin x}{|x|}$$

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(x^{2}-9\right) /(x+3).\) a. Make a table of the values of \(f\) at the points \(x=-3.1\) \(-3.01,-3.001,\) and so on as far as your calculator can go. Then estimate \(\lim _{x \rightarrow-3} f(x) .\) What estimate do you arrive at if you evaluate \(f\) at \(x=-2.9,-2.99,-2.999, \ldots\) instead? b. Support your conclusions in part (a) by graphing \(f\) near \(c=-3\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow-3.\) c. Find \(\lim _{x \rightarrow-3} f(x)\) algebraically, as in Example 7.

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=3 x-4, \quad x=2$$

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