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Given that \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x^{2}}=\frac{1}{2}, \quad\) quickly \(\quad\) evaluate \(\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1-\cos x}}{x}\)

Short Answer

Expert verified
The limit \(\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1-\cos x}}{x}\) equals \(+\infty\).

Step by step solution

01

Recognize the structure

Since we are supposed to find \(\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1-\cos x}}{x}\), we need to utilize the given hint \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x^{2}}=\frac{1}{2}\) by making the main question look like the hint.
02

Use the core property of limits

Recall the limit property that \(\lim _{x\rightarrow 0}(\sqrt{a}-\sqrt{b})=\frac{a-b}{\sqrt{a}+\sqrt{b}}\). We can apply this property directly by setting \(a=1\) and \(b=\cos{x}\), which transforms our target limit into: \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x (\sqrt{1}+\sqrt{cos(x)})}\)
03

Substitute the given limit

We can now notice that \(\frac{1-\cos x}{x^{2}}\) is in our equation. This allows us to use the given \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x^{2}}=\frac{1}{2}\) and substitute it giving \(\lim _{x \rightarrow 0^{+}} \frac{1}{2x (\sqrt{1}+\sqrt{cos(x)})}\)
04

Evaluate at 0

As \(x\) approaches 0, the \(\sqrt{\cos{x}}\) term also approaches 1 (since \(cos(0) = 1\)). This allows us to simplify the limit to \(\lim _{x \rightarrow 0^{+}} \frac{1}{2x(1+1)}\). Simplifying further gives \(\lim _{x \rightarrow 0^{+}} \frac{1}{4x}\)
05

Calculate the limit

The limit \(\lim _{x \rightarrow 0^{+}} \frac{1}{4x}\) is undefined, as division by zero is undefined. However, because we only consider positive \(x\) values (as shown by the \(+\) in the limit), the limit is \(+\infty\).

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Key Concepts

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

Trigonometric Limits
Trigonometric limits are a significant part of calculus, often involving ratios of trigonometric functions as the variable approaches a specific value. This exercise provides an excellent illustration of applying trigonometric identities to solve limits. Here, we deal with the expression \(1 - \cos x\) within a limit context. As \(x\) approaches 0, \(\cos x\) approaches \(1\), making \(1 - \cos x\) approach 0. This kind of behavior often leads to indeterminate forms, like \(\frac{0}{0}\), which require special techniques to evaluate. Utilizing given information, such as the limit \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x^{2}} = \frac{1}{2}\), helps transform complex expressions and leverage known results for simplification.
Calculus Problem Solving
Solving calculus problems often involves breaking down expressions into simpler components. In this problem, the approach is to identify the form of the equation and compare it to known limits. We start by transforming \(\frac{\sqrt{1-\cos x}}{x}\) into a recognizable form using algebraic manipulation and trigonometric identities. This requires using core calculus problem-solving skills, such as pattern recognition and substitution. The derived form \(\frac{1-\cos x}{x (\sqrt{1}+\sqrt{\cos x})}\), influenced by the original limit, sets the stage for further simplification. Identifying equivalent forms and making appropriate substitutions are often necessary in tackling calculus problems involving limits, especially where trigonometric functions and square roots are involved.
Limit Properties
Limit properties are essential tools in calculus problem-solving, allowing us to manipulate and evaluate limits effectively. One significant property used here is the limit of a function composed with roots: \(\lim _{x\rightarrow 0}(\sqrt{a}-\sqrt{b})=\frac{a-b}{\sqrt{a}+\sqrt{b}}\).This provides a way to transform and simplify the problem. In our exercise, choosing \(a = 1\) and \(b = \cos x\) allows us to express \(\sqrt{1-\cos x}\) concerning \(x\), simplifying its evaluation.Another core property is substituting available limit expressions, which leverages known solutions to reduce the complexity of the given problem. Utilizing properties effectively often involves recognizing equivalent fractions and applying substitutions where recognized limits have been previously evaluated.
Indeterminate Forms
Indeterminate forms arise when evaluating limits where direct substitution leads to expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not provide a finite or meaningfully interpretable result without further analysis.In this exercise, the expression \(\frac{\sqrt{1-\cos x}}{x}\) as \(x\) approaches 0 leads to an indeterminate form which requires resolving either through algebraic manipulation or using known limits. By using the hint \(\lim _{x \rightarrow 0^{+}} \frac{1-\cos x}{x^2} = \frac{1}{2}\), the problem leverages this data to bypass the indeterminacy and reach a solution. Indeterminate forms are a standard feature of trigonometric and calculus problems, necessitating a deep understanding and methodical approach to solve.

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Most popular questions from this chapter

A metal washer of (outer) radius \(r\) inches weighs \(2 r^{2}\) ounces. A company manufactures 2 -inch washers for different customers who have different error tolerances. If the customer demands a washer of weight \(8 \pm \varepsilon\) ounces, what is the error tolerance for the radius? That is, find \(\delta\) such that a radius of \(r\) within the interval \((2-\delta, 2+\delta)\) guarantees a weight within \((8-\varepsilon\) \(8+\varepsilon)\)

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