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Ushtrime t zgjidhura
Limitet dhe zbatimi i limiteve
Ass.Halil Snope
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LIMITT-T gjinden limitet e mposhtme:
) ( ) ( )
) ( ) ( ) ( ) ( )
)
) ( )
)
2 2
2 2
2 2
3 3
2 2
5
2 2
5 5
1 lim 3 5 2 lim 3 2 5 2 2 4
2 lim 1 1 lim 3 1 3 1 16
1 2 1 33 lim lim
2 2 2 4
34 lim 0 0 lim
5
3 10 5 2 105 lim lim l
5 5
x x
x x
x x
x
x x
x x
x x
x
x
xpasi emruesi tenton n dhe numruesi iti nuk ekziston
x
x x x x x
x x
+ = + =
+ = + =
+ += =
+ ++
+ = =
( ) ( )
( ) ( )( )
) ( ) ( )
( ) ( )( )
) ( ) ( )
( )
5
5 5
2 2
2 2 2
2 2
2 2
2 22 2 2
5 2 5im
5
2 5lim lim 2 7
5
2 3 26 2 3 66 lim lim lim
2 2 2
3 2lim lim 3 5
2
3 2 36 3 2 67 lim lim lim
23 2 2 2
x
x x
x x x
x x
x x x
x x x
x
x xx
x
x x xx x x x x
x x x
x xx
x
x x xx x x x x
x xx x x x x
+ =
+
= = + =
+ + + = = = =
+
= = + =
+ + = =
++ + + + +
( ) ( )
( ) ( )
)( ) ( )
( ) ( )( ) ( )
( ) ( )
( ) ( )
) ( )
( )( )
2 2
2 2
21 1 1
1 1
22
4 4 4
( 2)
2 3 3 5lim lim 5
1 2 1 1
1 5 14 5 5 58 lim lim lim
1 1 1 11
5 1 5 6lim lim 3
1 1 1 2
22 2 29 lim lim lim
4 4 2 4 2
x x
x x x
x x
x x x
x
x x x
x x x
x x xx x x x x
x x x xx
x x x
x x x
xx x x
x x x x x
=+ +
+ = = = =
+ + +
+ + + = = = + +
+ += = = =
+ +
+= = =
+ +
( )( )
) ( )
( )( )
( )( )
4 4
22
9 9 9
9 9
4 1 1lim lim
424 2
33 3 310 lim lim lim
9 9 3 9 3
9 1 1lim lim
639 3
x x
x x x
x x
x
xx x
xx x x
x x x x x
x
xx x
= = =
+ + +
= = = + +
= = =
+ +
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) ( )
( )( )
( )( ) ( )( )
) ( )
( )( )
22
3 3 3
3 3 3
22
5 5 5
5
1 21 2 1 2 1 211 lim lim lim
3 3 1 2 3 1 2
1 4 3 1 1lim lim lim
41 23 1 2 3 1 2
2 1 32 1 3 2 1 3 2 1 312 lim lim lim
5 5 2 1 3 5 2 1 3
2lim
x x x
x x x
x x x
x
xx x x
x x x x x
x x
xx x x x
xx x x
x x x x x
x
+ + + + += = =
+ + + +
+ = = = =
+ + + + + +
+= = =
+ +
=
( )( )( )
( )( )
) ( )
5 5
5
0
2 51 9 2lim lim
2 1 35 2 1 3 5 2 1 3
2 1lim 6 3
13 lim 2sin cos 2sin 0 cos0 2 0 1 1
x x
x
x
x
xx x x x
x x
= = =
+ + +
= =
+ = + = + =o o
)14 T c!tohet limiti i djtht dhe i mjt3 3
lim ( ) lim ( )x x
f x dhe f x +
22 " 3( )
3 " 3
x x pr xnse f x
x pr x