límits i continuïtat de funcions - matessantboianes

Solucionari
cos
cos cos
2
1− x ( 1− x)( 1+
x)
1−
cos x
c) lim = lim
= lim
=
x→02 x
x→02
x ( 1+
cos x ) x→0
2
x ( 1+
cos x)
2
sin x ⎛ 2 x 1 ⎞
= lim
= lim
⎜ ⋅
⎟
x→0
2 x x x 0 2
( 1+
cos ) ⎝⎜
x 1+
x ⎠
sin
1
=
→
cos ⎟ 2
cos
cos cos
2
1−x( 1− x)( 1+
x)
1−
cos x
d) lim = lim
= lim
=
x→0 x
x→0
x ( 1+
cos
x ) x→0
x( 1+
cos x )
2 sin x ⎛
= lim
= ⎜ sin x sin x ⎞
lim ⎜ ⋅ = 0
x→0
x ( 1+ cos x ) x→0⎝⎜
x 1+
cos x ⎠ ⎟
⎧⎪
2
⎪ x − 1 si x < 3
091 Si g( x)=
⎪
⎨
⎪
3
, determina els límits:
⎪ si x ≥ 3
⎩⎪
x + 5
a) lim
x →−1
b) lim
x →−5
g( x)
g( x)
c) lim g( x)
x → 3
d) lim g( x)
x → 6
2 a) lim g( x) = lim ( x − 1) = 0
x→−1 x→−1
2 b) lim g( x) = lim ( x − 1) = 24
c)
x→−5 x→−1
e) lim
x → +`
f) lim
x →−`
g( x)
g( x)
7
2
lim g( x) = lim ( x − 1) = 8
− −
x→3 x→3
3 3
→ lim g x lim
−
lim g( x)
= lim
x→3 x→3
+
+
x→3
x→3
x + 5 8
=
⎫ ⎪
⎬ ( ) ≠ g( x) → No existeix lim g( x).
⎪
+
x→3
⎪
⎭⎪
3 3
d) lim g( x)=
lim
x→6 x→6
x + 5 11
=
e) lim g( x)
= lim
x→+ ` x→+
`
x + =
3
0
5
f) lim g( x) = lim ( x − ) =+ `
x→−` x→−`
2 1
⎧ ⎪
4
⎪
si x 3
a) lim
x →−5
h( x)
b) lim h( x)
x → 2
c) lim h( x)
x → 5
d) lim
x →−2
h( x)
e) lim h( x)
x → 3
f) lim
x → +`
h( x)
445