limit as z approaches i of (z-i)/(z^2+1)

解答

z \to i lim (\frac{z - i}{z ^{2} + 1})

- \frac{i}{2}

求解步骤

z \to i lim (\frac{z - i}{z ^{2} + 1})

乘以

z - i

的共轭根式

: \frac{z ^{2} + 1}{z + i}

= z \to i lim (\frac{\frac{z ^{2} + 1}{z + i}}{z ^{2} + 1})

化简

\frac{\frac{z ^{2} + 1}{z + i}}{z ^{2} + 1} : \frac{1}{z + i}

= z \to i lim (\frac{1}{z + i})

代入值

z = i

= \frac{1}{i + i}

化简

\frac{1}{i + i} : - \frac{i}{2}

= - \frac{i}{2}

x \to \infty lim (x (\frac{π}{4} - arctan (\frac{x}{x + 1})))

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