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Evaluate the given limit without using L'Hôpital's rule, and then check that your answer is correct using L'Hôpital's rule.(a) $\lim _{x \rightarrow 0} \frac{\sin x}{\tan x}$(b) $\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1}$
(a) $\lim _{x \rightarrow 0} \frac{\sin x}{\tan x}=1$(b) $\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1}=\frac{2}{3}$
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
Campbell University
Baylor University
University of Michigan - Ann Arbor
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So in this problem, we're going Teoh evaluate limits without using low petals rule. So we have signed over tan and let's go ahead and simple and rewrite this using the definition of tan that sign over co sign. Now notice that the science cancel So we're just left with co sign co sign at zero is one. That's our answer now for Part two. The limit goes to one and looks like yet, or we were going to wants a factor, so we have X minus one X plus one on the bottom X minus one X squared plus X plus one and looks like the X minus ones canceled. Explosive one X squared plus X plus one on the bottom. That's two thirds that's it.
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