Problem 126
Question
\(\lim _{x \rightarrow \infty} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}\) is \(\quad\) (A) 1 (B) \(-1\) (C) Zero (D) Does not exist
Step-by-Step Solution
Verified Answer
The limit is zero.
1Step 1: Simplify the expression inside the limit
Begin by examining the expression \( \sqrt{1 - \cos 2x} \). Use the trigonometric identity \( \cos 2x = 1 - 2 \sin^2 x \) to rewrite \( \sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x| \). This simplifies our limit expression to \( \frac{\sqrt{2} |\sin x|}{\sqrt{2} x} \).
2Step 2: Simplify the fraction
Cancel out the common factor \( \sqrt{2} \) in the numerator and denominator. Our expression now reduces to \( \frac{ |\sin x| }{ x } \).
3Step 3: Evaluate the limit
Examine the behavior of \( \frac{ |\sin x| }{ x } \) as \( x \rightarrow \infty \). The function \( \sin x \) oscillates between -1 and 1, so \( |\sin x| \) is always between 0 and 1. Thus, as \( x \rightarrow \infty \), the fraction \( \frac{|\sin x|}{x} \rightarrow 0 \) because the numerator is bounded while the denominator grows without bound.
Key Concepts
Trigonometric IdentitiesInfinite LimitsOscillating Functions
Trigonometric Identities
Trigonometric identities are powerful tools that help simplify expressions involving trigonometric functions. One of these essential identities is the double-angle cosine identity, which states that \( \cos 2x = 1 - 2 \sin^2 x \). This identity allows us to express cosine in terms of sine, facilitating simplification in more complex problems.
In the original exercise, we start with the expression \( \sqrt{1 - \cos 2x} \). By substituting the identity, we transform it to \( \sqrt{1 - (1 - 2 \sin^2 x)} = \sqrt{2 \sin^2 x} \). This further simplifies to \( \sqrt{2} |\sin x| \) due to the property of square roots and absolute values.
This transformation illustrates the importance of understanding and applying trigonometric identities, especially when dealing with expressions that seem challenging at first glance. Utilizing identities can often turn a complex problem into one that is much more manageable by revealing the underlying structure of the trigonometric components.
In the original exercise, we start with the expression \( \sqrt{1 - \cos 2x} \). By substituting the identity, we transform it to \( \sqrt{1 - (1 - 2 \sin^2 x)} = \sqrt{2 \sin^2 x} \). This further simplifies to \( \sqrt{2} |\sin x| \) due to the property of square roots and absolute values.
This transformation illustrates the importance of understanding and applying trigonometric identities, especially when dealing with expressions that seem challenging at first glance. Utilizing identities can often turn a complex problem into one that is much more manageable by revealing the underlying structure of the trigonometric components.
Infinite Limits
Infinite limits in calculus help us understand the behavior of functions as the input variable grows very large, approaching infinity. When we evaluate the limit \( \lim_{x \rightarrow \infty} f(x) \), we seek to determine what value, if any, the function \( f(x) \) approaches as \( x \) becomes infinitely large.
In our exercise, the expression we are interested in is \( \frac{ |\sin x| }{ x } \) as \( x \rightarrow \infty \). The numerator, \( |\sin x| \), oscillates between 0 and 1. This implies it remains bounded regardless of how large \( x \) gets. Meanwhile, the denominator \( x \) grows without bound, becoming infinitely large as \( x \rightarrow \infty \).
Since the numerator remains small (and is, in fact, always less than or equal to 1), and the denominator becomes increasingly large, the fraction approaches 0. This means that \( \lim_{x \rightarrow \infty} \frac{ |\sin x| }{ x } = 0 \). Understanding how expressions behave at infinity is crucial, particularly in calculus, where limits provide insight into rates of growth and the convergence of sequences and series.
In our exercise, the expression we are interested in is \( \frac{ |\sin x| }{ x } \) as \( x \rightarrow \infty \). The numerator, \( |\sin x| \), oscillates between 0 and 1. This implies it remains bounded regardless of how large \( x \) gets. Meanwhile, the denominator \( x \) grows without bound, becoming infinitely large as \( x \rightarrow \infty \).
Since the numerator remains small (and is, in fact, always less than or equal to 1), and the denominator becomes increasingly large, the fraction approaches 0. This means that \( \lim_{x \rightarrow \infty} \frac{ |\sin x| }{ x } = 0 \). Understanding how expressions behave at infinity is crucial, particularly in calculus, where limits provide insight into rates of growth and the convergence of sequences and series.
Oscillating Functions
Oscillating functions, such as trigonometric functions like sine and cosine, are notable for their periodic and repeated patterns over specific intervals. Due to this behavior, these functions do not settle at a single value but continuously fluctuate within a defined range.
The sine function, \( \sin x \), oscillates between -1 and 1. Consequently, the absolute value, \( |\sin x| \), oscillates between 0 and 1, making it a bounded function. Despite \( \sin x \)'s constant change in direction and magnitude, its absolute value maintains a consistent range.
In the exercise at hand, understanding the nature of \( |\sin x| \) is vital. While \( x \) increases and tends towards infinity, \( |\sin x| \)'s oscillating behavior results in it remaining within the fixed interval \([0, 1]\). The key takeaway is that while oscillating functions like sine and cosine continually change, their bounded nature often simplifies the evaluation of limits, particularly as variables approach infinity. This example underlines the fundamental concept of how oscillatory patterns, combined with large denominators, can drive expressions toward a limiting value like 0.
The sine function, \( \sin x \), oscillates between -1 and 1. Consequently, the absolute value, \( |\sin x| \), oscillates between 0 and 1, making it a bounded function. Despite \( \sin x \)'s constant change in direction and magnitude, its absolute value maintains a consistent range.
In the exercise at hand, understanding the nature of \( |\sin x| \) is vital. While \( x \) increases and tends towards infinity, \( |\sin x| \)'s oscillating behavior results in it remaining within the fixed interval \([0, 1]\). The key takeaway is that while oscillating functions like sine and cosine continually change, their bounded nature often simplifies the evaluation of limits, particularly as variables approach infinity. This example underlines the fundamental concept of how oscillatory patterns, combined with large denominators, can drive expressions toward a limiting value like 0.
Other exercises in this chapter
Problem 124
In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True
View solution Problem 125
In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True
View solution Problem 127
\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+2}\right)^{x}\) is equal to (A) \(e^{4}\) (B) \(e^{2}\) (C) \(e^{3}\) (D) \(e\)
View solution Problem 129
Let \(f(2)=4\) and \(f^{\prime}(2)=4\). Then \(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\) is given by (A) 2 (B) \(-2\) (C) \(-4\) (D) 3
View solution