$\lim _{x \rightarrow 0}\left(\frac{2 x^2-x \cdot \sin (x)}{x \cdot \ln (x)-2 \sin ^2(x)}\right)$
Answer (2)
The limit is evaluated as lim x → 0 ( x ⋅ l n ( x ) − 2 s i n 2 ( x ) 2 x 2 − x ⋅ s i n ( x ) ) = 0 , as it simplifies to an indeterminate form of type 0 0 . We applied L'Hopital's rule twice to analyze the behavior of the numerator and denominator, confirming that the final limit is 0. Therefore, the answer is 0 .
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Recognize the limit is in the indeterminate form 0 0 .
Apply L'Hopital's rule by differentiating the numerator and the denominator.
Apply L'Hopital's rule a second time to simplify the expression further.
Evaluate the limit as x approaches 0, resulting in 0 .
Explanation
Problem Analysis We are asked to find the limit of the expression x ⋅ l n ( x ) − 2 s i n 2 ( x ) 2 x 2 − x ⋅ s i n ( x ) as x approaches 0.
Indeterminate Form The limit is of the indeterminate form 0 0 as x → 0 , since 2 ( 0 ) 2 − 0 ⋅ sin ( 0 ) = 0 and 0 ⋅ ln ( 0 ) − 2 sin 2 ( 0 ) = 0 . We can apply L'Hopital's rule to evaluate the limit.
First Application of L'Hopital's Rule Applying L'Hopital's rule involves taking the derivative of the numerator and the derivative of the denominator with respect to x . The derivative of the numerator 2 x 2 − x sin ( x ) is 4 x − sin ( x ) − x cos ( x ) . The derivative of the denominator x ln ( x ) − 2 sin 2 ( x ) is ln ( x ) + 1 − 4 sin ( x ) cos ( x ) = ln ( x ) + 1 − 2 sin ( 2 x ) .
Analyzing the New Limit The new limit is lim x → 0 l n ( x ) + 1 − 2 s i n ( 2 x ) 4 x − s i n ( x ) − x c o s ( x ) . As x approaches 0, the numerator approaches 0, and the denominator approaches − ∞ . Thus, the limit is of the form − ∞ 0 , which suggests the limit is 0.
Second Application of L'Hopital's Rule To confirm, let's apply L'Hopital's rule again. The derivative of 4 x − sin ( x ) − x cos ( x ) is 4 − cos ( x ) − cos ( x ) + x sin ( x ) = 4 − 2 cos ( x ) + x sin ( x ) . The derivative of ln ( x ) + 1 − 2 sin ( 2 x ) is x 1 − 4 cos ( 2 x ) .
Rewriting the Limit The new limit is lim x → 0 x 1 − 4 c o s ( 2 x ) 4 − 2 c o s ( x ) + x s i n ( x ) . This can be rewritten as lim x → 0 1 − 4 x c o s ( 2 x ) x ( 4 − 2 c o s ( x ) + x s i n ( x )) .
Final Evaluation Now, as x → 0 , the numerator approaches 0 ( 4 − 2 ( 1 ) + 0 ) = 0 and the denominator approaches 1 − 4 ( 0 ) ( 1 ) = 1 . Therefore, the limit is 1 0 = 0 .
Conclusion Thus, lim x → 0 ( x ⋅ l n ( x ) − 2 s i n 2 ( x ) 2 x 2 − x ⋅ s i n ( x ) ) = 0 .
Examples
In physics, when analyzing the behavior of systems near equilibrium, you might encounter limits similar to this one. For example, when studying damped oscillations, the ratio of energy dissipation to the frequency of oscillation can be expressed as a limit. Evaluating such limits helps physicists understand the stability and long-term behavior of these systems. Understanding how to manipulate and evaluate limits using L'Hopital's rule is crucial for making accurate predictions about physical phenomena.
