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Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy


Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy

Using the squeeze theorem to prove that the limit as x approaches 0 of (sin x)/x =1 Watch the next lesson: https://www.khanacademy.org/math/differential-calculus/limits_topic/epsilon_delta/v/limit-intuition-review?utm_source=YT&utm_medium=Desc&utm_campaign=DifferentialCalculus Missed the previous lesson? https://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/squeeze-theorem?utm_source=YT&utm_medium=Desc&utm_campaign=DifferentialCalculus Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to KhanAcademy’s Differential Calculus channel: https://www.youtube.com/channel/UCNLzjGl1HBdZrHXo4Vae3iA?sub_confirmation=1 Subscribe to KhanAcademy: https://www.youtube.com/subscription_center?add_user=khanacademy
動画ID:Ve99biD1KtA
投稿日時:2008年06月25日 12時58分
再生回数:1,034,040 回
コメント数:468
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