2.Limity-příklady
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x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
We have to convert the function inside the limit into fraction. We use
the identity
e−
x
2
=
1
ex
2
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
2e−x
2
lim
x→∞
x
2e−x
2
=
∞ × 0
pr.
= lim
x→∞
x
2
ex
2
=
∞
∞
l0H.
= lim
x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
The limit has the form required by the l’ Hospital rule.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
2e−x
2
lim
x→∞
x
2e−x
2
=
∞ × 0
pr.
= lim
x→∞
x
2
ex
2
=
∞
∞
l0H.
= lim
x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
We use l’ Hospital rule.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
2e−x
2
lim
x→∞
x
2e−x
2
=
∞ × 0
pr.
= lim
x→∞
x
2
ex
2
=
∞
∞
l0H.
= lim
x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
We simplify.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
2e−x
2
lim
x→∞
x
2e−x
2
=
∞ × 0
pr.
= lim
x→∞
x
2
ex
2
=
∞
∞
l0H.
= lim
x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
Dosadı´me (in the sense of limits). Hence we evaluate separately
the limit of the numerator and the denominator.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
2e−x
2
lim
x→∞
x
2e−x
2
=
∞ × 0
pr.
= lim
x→∞
x
2
ex
2
=
∞
∞
l0H.
= lim
x→∞
2x
2xex
2
pr.
= lim
x→∞
1
ex
2
=
1
∞
= 0
We obtain well–defined expression which equals zero. The problem
is solved.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
(arctg x −
π
2
)
lim
x→∞
x
arctg x −
π
2
=
∞ × 0
pr.
= lim
x→∞
arctg x −
π
2
1
x
=
0
0
l0H.
= lim
x→∞
1
1 + x2
−
1
x2
=
0
0
pr.
= lim
x→∞
−x
2
1 + x2
=
−∞
∞
= lim
x→∞
−x
2
x2
=
−1
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
(arctg x −
π
2
)
lim
x→∞
x
arctg x −
π
2
=
∞ × 0
pr.
= lim
x→∞
arctg x −
π
2
1
x
=
0
0
l0H.
= lim
x→∞
1
1 + x2
−
1
x2
=
0
0
pr.
= lim
x→∞
−x
2
1 + x2
=
−∞
∞
= lim
x→∞
−x
2
x2
=
−1
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→∞
x
(arctg x −
π
2
)
lim
x→∞
x
arctg x −
π
2
=
∞ × 0
pr.
= lim
x→∞
arctg x −
π
2
1
x
=
0
0
l0H.
= lim
x→∞
1
1 + x2
−
1
x2
=
0
0
pr.
= lim