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7/26/2019 Gua N 1 Limite de Funciones
1/2
Gua N 1: Limites de funciones
Docente : Auristela Olivares Rojas
Asignatura : Clculo
1.
Aplique operatoria algebraica convencional y verifique los resultados;
1. 11
21
=
xlimx
1.3x
lim3
2142
+
x
xx! 1"
#.2x
lim( 654 2 = xx 1$.0x
lim22
22
2 aaxx
ax
++
! %1&
a "
. ( ) 7104
2
1
=+
xxlimx 1'. 0a
lim22
22
2 aaxx
ax
++
! 1& (
"
$.0n
lim63
53
2
2
+
+
nn
nn!6
51).
axlim 22
22
2 aaxx
ax
++
! "& a "
'.3x
lim3
2142
+
x
xx! 1" 1*.
0xlim
275
225
7
3
+
x
x! %1
).2x
lim32
56
2
2
++
xx
xx! %* 1+. 4
2
42
2
=
x
xlimx
*.1x
lim
32
56
2
2
++
xx
xx! %1 1,.
2xlim
2
22
++
x
x!2
3
+.1x
lim3
92
x
x! $ #".
1x
lim
1
12
x
x! "
,.3x
lim2
3
2
x
x! 0 #1.
2x
lim
2
83
++
x
x! 1#
1".2y
lim2
65
2
+
++
yyy ! ' ##.
4xlim
4
643
xx ! $+
11.2y
lim
2
652
+
y
yy! %1 #.
2xlim
253
103
2
2
+xx
xx! 1
1#.3x
lim
3
342
+++
x
xx! %#
#$.2h
lim6
44
2
23
++
hh
hhh!
#'.ax
lim 22
33
xa
ax
! %
2
3 a
#).4x
limxx
xxx
4
433
2
23
!4
21
7/26/2019 Gua N 1 Limite de Funciones
2/2
#*.1x
lim1
362 234
++
x
xxx! %+
#+.1h
lim ( )
h
h 11 2 +
!
#,.1x
limx
x
2
33 ! "
".2x
lim
2
2
11
x
x ! %4
1
1.2x
lim8
16
3
4
x
x!3
8
#.1x
lim
9
3
3
9
2
2
x
x
x
x!
4
15
.3
4
2
1
2
23
1
=
+
+
xx
xxxlimx
$.3
2
2
1
2
23
1
=
+
+
xx
xxxlimx
'.3
4
1
32
3
2
1
=+
x
xxlimx