Chapter 4: Problem 49
Predict the value of the limit. Then find each limit using I'Hopital's rule. $$\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{x}$$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Analyze the Limit's Form
First, evaluate the limit \( \lim_{x \to 0} \frac{\sin^2 x}{x} \) by substituting \( x = 0 \) and see if it results in an indeterminate form. Substituting directly gives \( \frac{0^2}{0} = \frac{0}{0} \), which is indeterminate.
02
Predict the Limit Value
Because as \( x \to 0 \), \( \sin x \to x \), we can predict that \( \sin^2 x \to x^2 \). Hence, the limit resembles \( \frac{x^2}{x} \to x \), which suggests the limit is 0.
03
Apply L'Hopital's Rule
L'Hopital's Rule is applicable for \( \frac{0}{0} \) indeterminate forms. Differentiate the numerator and denominator: \( f(x) = \sin^2 x \Rightarrow f'(x) = 2\sin x \cos x \) \( g(x) = x \Rightarrow g'(x) = 1 \). Thus, the limit becomes: \[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2 \sin x \cos x}{1} = 2 \lim_{x \to 0} \sin x \cos x \].
04
Evaluate the New Limit
Evaluate \( \lim_{x \to 0} 2 \sin x \cos x \): \( \sin x \to 0 \) and \( \cos x \to 1 \) as \( x \to 0 \). Therefore, the limit becomes \( 2 \cdot 0 \cdot 1 = 0 \).
05
Conclude the Result
The value of the limit \( \lim _{x \to 0} \frac{\sin ^{2} x}{x} \) has been found using L'Hopital's Rule and equals 0, which confirms the earlier prediction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
Indeterminate forms occur in calculus when evaluating a limit leads to expressions that are not defined in a straightforward manner. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \). When you encounter such forms, it's essential to employ methods like L'Hopital's Rule to resolve them effectively. In the exercise, substituting \( x = 0 \) in \( \lim_{x \to 0} \frac{\sin^2 x}{x} \) initially results in the indeterminate form \( \frac{0}{0} \). This type of indeterminate form signals that further analysis or alternate techniques are needed to find the actual limit. It is a crucial step in ensuring the solution represents a meaningful outcome rather than a mathematically undefined result.
Trigonometric Limits
Trigonometric limits are limits that involve trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \). These often come up in calculus due to their unique properties near zero. Particularly, when \( x \to 0 \), \( \sin x \approx x \) and \( \cos x \approx 1 \). Knowing these approximations allows predictions about the behavior of trigonometric functions as they approach zero. In our specific exercise, because \( \sin x \to x \) as \( x \to 0 \), it suggests that \( \sin^2 x \approaches x^2 \). Understanding these properties is invaluable when addressing limits, enabling one to conceptualize how trigonometric functions simplify when approaching certain values.
Differentiation
Differentiation is a fundamental principle in calculus that provides the rate at which a function changes at any given point. To apply L'Hopital's Rule, one needs to differentiate both the numerator and the denominator of a function separately. For our problem, the function \( f(x) = \sin^2 x \) was differentiated resulting in \( f'(x) = 2\sin x \cos x \). The linear denominator \( g(x) = x \) differentiated to \( g'(x) = 1 \). This transformation allowed the initial indeterminate form \( \frac{0}{0} \) to be resolved. Differentiation translates a complex form into something manageable, where the limit of \( \frac{f'(x)}{g'(x)} \) can be directly evaluated as \( 2 \lim_{x \to 0} \sin x \cos x \). Mastery of differentiation not only aids in resolving limits with L'Hopital's Rule but is also pivotal across all branches of calculus.