ściaga Analiza 1a.pdf
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2.ℎśóął|lim
→
|
ż|3...|: ∈;ąłóż|
= lim
→
śćąℎą|
|
> 0 ∈ ∩
ą ∈ |ą ∈ | = .
= .. = .. ≤ lim
. =
.. ≥ lim
|4.ł|. = …| = …
|
=
ąłł
3
+1
=
+
+
+1
|.|ł.óń
.
,,| =
+
+
+1
+| = .|5.ł| ≤ ≤ | ≤ ≤ | =
|5.| ≤ ≤ | = …| =
1+
|′ = |
5.| ≤ ≤ | ≤ ≤ | =
−
|6.łłś|
| =
=
= ′ =
′ = =
= lim
→
|
../.| =
|
}=
= [′]
−
= |.|lim
→
|. = lim
→
|3.
.ś|: ∈ ;| lim
→
| lim
→
|.ś| = +
|żś
|
= |
> 0||ℎℎ|
= lim
→
′ = 0|′ = |
′=
|
′= −
|
√
′=
1
2
√
|
=
|
′=
ln|′ =
1
|
1
ln
|sin′= |′ = −|′ =
1
√
1−
−1
√
1−
log
′ =
|′ =
|
1
= 1+
|′ =
−1
1
1+
|′ =
−1
′ =
1+
łℎ|+′= ′+′|−′= ′−′|′ = ′|′ = ′+′|
= −1+
|′=
=
−
[]′= ′′|ł|
=
+1
+|
=
|+|
=
+|
=
ln
+|
= − +|
= +|
= −ln||+
| = ln||+
= +
= |= ′|.|||łęś|
=
= − +|ł.
= [′−
|
=
1−2
2
|
=
1+2
2
|2 = 2|
2
2+1
= 1+
2
2+1
=
2 =
−
|1+
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