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
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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.
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Видео Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy канала Khan Academy
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
Видео Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy канала Khan Academy
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