Limit of (arcsen(2x)-2arcsen(x))/x^3 as x->0

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SUMMARY

The limit of the expression (arcsen(2x) - 2arcsen(x)) / x^3 as x approaches 0 is evaluated using L'Hôpital's rule due to the 0/0 indeterminacy. The derivative of arcsen(x), given by 1/Sqrt(1-x^2), is crucial in simplifying the expression. The discussion highlights the necessity of applying L'Hôpital's rule and suggests that alternative methods may not effectively eliminate the trigonometric functions involved.

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  • Understanding of limits and indeterminate forms in calculus
  • Familiarity with L'Hôpital's rule for resolving limits
  • Knowledge of the derivative of the arcsine function, arcsen(x)
  • Basic algebraic manipulation skills, including working with conjugates
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  • Study advanced applications of L'Hôpital's rule in calculus
  • Explore Taylor series expansions for trigonometric functions
  • Learn about alternative limit evaluation techniques, such as the epsilon-delta definition
  • Investigate the behavior of arcsine and its derivatives near zero
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Students studying calculus, particularly those focusing on limits and derivatives, as well as educators seeking to enhance their teaching methods for resolving indeterminate forms.

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Homework Statement


Limit of (arcsen(2x)-2arcsen(x))/x^3 as x->0


Homework Equations


arsen(x)'=1/Sqrt(1-x^2)


The Attempt at a Solution


It's an 0/0 indeterminancy, to solve it, I had to use L'hopital's rule once simplify de expression, multiplying by the conjugate root. Is there an easier way?
 
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As you don't get rid of the trigonometric functions without l'Hospital, I think this ok.

$$\frac{\frac{2}{\sqrt{1-(2x)^2}}-\frac{2}{\sqrt{1-(x)^2}}}{x^2} = 2 \frac{1}{\frac{1}{\sqrt{1-(2x)^2}}+\frac{1}{\sqrt{1-(x)^2}}} \frac{\frac{1}{1-(2x)^2}-\frac{1}{1-(x)^2}}{x^2} \to \frac{2}{2} \frac{\frac{1}{1-(2x)^2}-\frac{1}{1-(x)^2}}{x^2} \to \dots$$
 
Thanks!
 

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