Evaluate [tex]$\lim _{x \rightarrow 0} \frac{x}{\sin ^{-1}(3 x)}=$[/tex]
Answer (2)
The limit lim x → 0 s i n − 1 ( 3 x ) x evaluates to 3 1 through the application of L'Hopital's Rule after recognizing an indeterminate form. Differentiating the numerator and denominator leads us to a determinate limit that simplifies the computation. Ultimately, substituting x = 0 gives us the final result of 3 1 .
;
Recognize the limit as an indeterminate form 0 0 .
Apply L'Hopital's Rule by differentiating the numerator and denominator.
Find the derivatives: d x d ( x ) = 1 and d x d ( sin − 1 ( 3 x )) = 1 − 9 x 2 3 .
Evaluate the limit: lim x → 0 3 1 − 9 x 2 = 3 1 .
3 1
Explanation
Problem Analysis We are asked to evaluate the limit of the function s i n − 1 ( 3 x ) x as x approaches 0 . This is a limit problem that can be solved using L'Hopital's Rule or by recognizing a standard limit.
Indeterminate Form As x approaches 0 , both the numerator x and the denominator sin − 1 ( 3 x ) approach 0 . This results in an indeterminate form of type 0 0 .
L'Hopital's Rule Since we have an indeterminate form of type 0 0 , we can apply L'Hopital's Rule, which states that if lim x → c g ( x ) f ( x ) is of the form 0 0 or ∞ ∞ , then lim x → c g ( x ) f ( x ) = lim x → c g ′ ( x ) f ′ ( x ) , provided the limit exists.
Finding Derivatives Let f ( x ) = x and g ( x ) = sin − 1 ( 3 x ) . Then, we need to find the derivatives of f ( x ) and g ( x ) with respect to x .
The derivative of f ( x ) = x is f ′ ( x ) = 1 .
The derivative of g ( x ) = sin − 1 ( 3 x ) is g ′ ( x ) = 1 − ( 3 x ) 2 1 ⋅ 3 = 1 − 9 x 2 3 .
Applying L'Hopital's Rule Now, we apply L'Hopital's Rule: x → 0 lim sin − 1 ( 3 x ) x = x → 0 lim 1 − 9 x 2 3 1 = x → 0 lim 3 1 − 9 x 2
Evaluating the Limit Now, we evaluate the limit by substituting x = 0 into the expression: x → 0 lim 3 1 − 9 x 2 = 3 1 − 9 ( 0 ) 2 = 3 1 = 3 1
Final Answer Therefore, the limit is 3 1 .
Examples
In physics, when analyzing the motion of a pendulum with a small angle of displacement, the approximation sin ( θ ) ≈ θ is often used. This approximation is based on the limit lim θ → 0 θ s i n ( θ ) = 1 . Similarly, the limit we evaluated, lim x → 0 s i n − 1 ( 3 x ) x = 3 1 , can be used in situations where inverse trigonometric functions appear in physical models, such as in optics or mechanics, when dealing with small angles or displacements. Understanding such limits helps simplify complex equations and provides accurate approximations for real-world phenomena.
