Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy
18:05
YouTubeKhan 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 ...
1.3M viewsJun 25, 2008
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