limit as x approaches 2pi of 1/(sin(x))

解答

x \to 2 π lim (\frac{1}{sin ( x )})

发散

求解步骤

x \to 2 π lim (\frac{1}{sin ( x )})

若

x \to a^{-} lim f (x) \neq = x \to a^{+} lim f (x)

则极限不存在

x \to 2 π + lim (\frac{1}{sin ( x )}) = \infty

x \to 2 π - lim (\frac{1}{sin ( x )}) = - \infty

= 发散

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