r ( t ) = { ( t , t 2 ) if t ∈ [ − 2 , 0 ] ( t , t ) if t ∈ ( 0 , 2 ) ( t , t 2 ) if t ∈ [ 2 , 3 ] r(t)= \begin{cases}
(t, t^2) &\text{if } t\in [-2,0] \\
(t, t) &\text{if } t\in (0,2) \\
(t, t^2) &\text{if } t\in [2,3] \\
\end{cases} r ( t ) = ⎩ ⎨ ⎧ ( t , t 2 ) ( t , t ) ( t , t 2 ) if t ∈ [ − 2 , 0 ] if t ∈ ( 0 , 2 ) if t ∈ [ 2 , 3 ]
(a) Domain: [ − 2 , 3 ] [-2, 3] [ − 2 , 3 ]
(b)
r ( 0 ) = ( 0 , 0 ) r(0)=(0, 0) r ( 0 ) = ( 0 , 0 )
lim t → 0 − r ( t ) = ( 0 , 0 ) \lim\limits_{t\to 0^-}r(t)=(0,0) t → 0 − lim r ( t ) = ( 0 , 0 )
lim t → 0 + r ( t ) = ( 0 , 0 ) \lim\limits_{t\to 0^+}r(t)=(0,0) t → 0 + lim r ( t ) = ( 0 , 0 ) Then
lim t → 0 r ( t ) = ( 0 , 0 ) = r ( 0 ) \lim\limits_{t\to 0}r(t)=(0,0)=r(0) t → 0 lim r ( t ) = ( 0 , 0 ) = r ( 0 ) The function r ( t ) r(t) r ( t ) is continuous at t = 0. t=0. t = 0.
(c)
r ( 2 ) = ( 2 , 4 ) r(2)=(2, 4) r ( 2 ) = ( 2 , 4 )
lim t → 2 − r ( t ) = ( 2 , 2 ) \lim\limits_{t\to 2^-}r(t)=(2,2) t → 2 − lim r ( t ) = ( 2 , 2 )
lim t → 2 + r ( t ) = ( 2 , 4 ) \lim\limits_{t\to 2^+}r(t)=(2,4) t → 2 + lim r ( t ) = ( 2 , 4 )
lim t → 2 − r ( t ) ≠ lim t → 2 + r ( t ) \lim\limits_{t\to 2^-}r(t)\not=\lim\limits_{t\to 2^+}r(t) t → 2 − lim r ( t ) = t → 2 + lim r ( t ) Then
lim t → 2 r ( t ) = does not exist \lim\limits_{t\to 2}r(t)=\text{does not exist} t → 2 lim r ( t ) = does not exist The function r ( t ) r(t) r ( t ) is not continuous at t = 2. t=2. t = 2.
(d)
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