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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