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