limit as t approaches 0+of cos(t)

解答

t \to 0 + lim (cos (t))

1

求解步骤

t \to 0 + lim (cos (t))

代入值

t = 0

= cos (0)

使用以下普通恒等式:

cos (0) = 1

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

h \to 0 lim (\frac{( x + h ) ^{3} - 3}{h})

x \to \infty lim (\frac{u ( x ) k + 1}{u ( x ) k})

x \to \frac{π}{2} lim (ln ∣ cos (x) ∣)

x \to 4 lim (\frac{2 - 7 x}{4 - x})

x \to 0 lim (\frac{2 ^{x} - 1}{2 x})