Q23·Section II (Mathematics)·23 of 84∣sinx−siny∣≤∣x−y∣|\sin x-\sin y|\le|x-y|∣sinx−siny∣≤∣x−y∣: Lipschitz, hence uniformly continuousf(x)=sinxf(x)=\sin xf(x)=sinx, x∈Rx\in\mathbb{R}x∈R: OptionsAf(x)f(x)f(x) is continuous but not uniformly continuousBf(x)f(x)f(x) is uniformly continuous but not LipschitzCf(x)f(x)f(x) is Lipschitz and uniformly continuousDf(x)f(x)f(x) is neither Lipschitz nor uniformly continuousPick an option to check your answer.Show answerShow solution Report a problem with this question← Q22All questionsQ24 →Keyboard: A–D answer · S solution · ←/→ previous/next