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Find the limits.$$\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x}{x}$$
$$2$$
Calculus 1 / AB
Chapter 3
TOPICS IN DIFFERENTIATION
Section 6
L Hopital's Rule; Indeterminate Forms
Functions
Limits
Derivatives
Differentiation
Continuous Functions
Applications of the Derivative
Missouri State University
Baylor University
University of Nottingham
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here. We're starting with limit as X goes to zero of sine inverse two X divided by X. Now notice that if we just try to do this limit, we get zero on the top and zero on the bottom. 00 is an indeterminant form. And so we're good to go to apply low petals rule. I'm writing LH over this equal sign indicating that that's next step. So the derivative of the numerator is going to be one over square root one minus two x squared and then we need to multiply by the derivative of two X is a chain room. Now the derivative, the denominator is just one. Now this new limit is something that we can easily dio Hogan zero on, then, just a value. It's 22 that's it.
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