2.Limity-příklady
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1
1
x
and
multiply with logarithm.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→0+
x
ln x.
lim
x→0+
x
ln x = 0 × (−∞)
pr.
= lim
x→0+
ln x
1
x
=
−∞
∞
l0H.
= lim
x→0+
1
x
−
1
x2
pr.
= lim
x→0+
−x = 0
Now we have indeterminate form for which l’ Hospital rule can be
used.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→0+
x
ln x.
lim
x→0+
x
ln x = 0 × (−∞)
pr.
= lim
x→0+
ln x
1
x
=
−∞
∞
l0H.
= lim
x→0+
1
x
−
1
x2
pr.
= lim
x→0+
−x = 0
We use l’ Hospital rule.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→0+
x
ln x.
lim
x→0+
x
ln x = 0 × (−∞)
pr.
= lim
x→0+
ln x
1
x
=
−∞
∞
l0H.
= lim
x→0+
1
x
−
1
x2
pr.
= lim
x→0+
−x = 0
We simplify. The function in the limit is continuous at x = 0.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→0+
x
ln x.
lim
x→0+
x
ln x = 0 × (−∞)
pr.
= lim
x→0+
ln x
1
x
=
−∞
∞
l0H.
= lim
x→0+
1
x
−
1
x2
pr.
= lim
x→0+
−x = 0
The limit of continuous function can be evaluated by direct
substitution.
//
/
.
..
c
Robert Marˇı´k, 2008 ×
Find
lim
x→0+
x
ln x.
lim
x→0+
x
ln x = 0 × (−∞)
pr.
= lim
x→0+
ln x
1
x
=
−∞
∞
l0H.
= lim
x→0+
1
x
−
1
x2
pr.
= lim
x→0+
−x = 0
The problem is solved.
//
/
.
..
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
//
/
.
..
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 start with the limit.
//
/
.
..
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. More precisely, we evaluate separately the following
limits
lim
x→∞
x2
and
lim
x→∞
e−
x
2
.
We obtain an indeterminate form.
//
/
.
..
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