AM1
Mathematical Analysis 1
Oct. 2011 – Feb. 2012
Exercises Lecture 13
Date: January 13
Example 13.1.
1.
1
lim 4 x
2.
x→0+
1
log 4
x
= +∞
1
log 4
x
=0
1. limx→0+ 4 x = limx→0+ e
2. limx→0− 4 x = limx→0− e
1
lim 4 x
x→0−
Example 13.2.
1
lim (sin x) log2 x
x→1+
1
lim (sin x) log2 x = lim e
log sin x
log2 x
x→1+
x→1+
=0
Since log sin x < 0 for any x such that sin x > 0.
Example 13.3.
lim |x − 1|x−1
x→1
lim |x − 1|x−1 = lim e(x−1) log |x−1| = lim e y log |y| = 1
x→1
x→1
y→0
Since limy→0 y log y = 0.
Example 13.4.
√
sin x − x
√
lim
x→0 1 − cos 4 x
We have that
sin x −
√
√
sin x
x( √ − 1)
x
√
= − x(1 + o(1))
x=
13-1
for x → 0
13-2
and
1 − cos
Thus
√
4
x=
1√
x(1 + o(1))
2
for x → 0
√
sin x − x
√ = −2
lim
x→0 1 − cos 4 x
Example 13.5.
lim x [log(x + 1) − log x]
x→+∞
We have that
x+1
lim x [log(x + 1) − log x] = lim x log
x→+∞
x→+∞
x
1
= lim x log 1 +
x→+∞
x
1
= lim log(1 + y)
y→0 y
y(1 + o(1))
= lim
y→0
y
=1
Example 13.6.
r
lim (x + 5)
x→+∞
x+1
−x
x−1
We have that
r
lim (x + 5)
x→+∞
Where
x+1
x−1
=1+
2
x−1
#
r
x+1
x+1
−1 +5
x−1
x−1
"r
#
2
= 5 + lim x
1+
−1
x→+∞
x−1
1 2
1
= 5 + lim x 1 +
+o
−1
x→+∞
2 x−1
x
x
= 5 + lim
x→+∞ x − 1
=6
x+1
− x = lim x
x→+∞
x−1
"r
is obtained by polynomial division.
13-3
Example 13.7.
lim
x→+∞
x3 − 2x + 1
x2 + x3
2 +1
2xx−3
First of all we rewrite the limit in the following equivalent form
lim e
2x2 +1
x−3
log
x3 −2x+1
x2 +x3
x→+∞
Next we perform the division of x3 − 2x + 1 by x3 + x2 and find
−x2 − 2x + 1
x3 − 2x + 1
=
1
+
x2 + x3
x3 + x2
Therefore
2x2 + 1
log
x−3
x3 − 2x + 1
x2 + x3
2x2 + 1
−x2 − 2x + 1
log 1 +
x−3
x3 + x2
2
2
−x − 2x + 1
2x + 1
(1 + o(1))
=
x−3
x3 + x2
−2x4 (1 + o(1))
=
(1 + o(1))
x4 (1 + o(1))
= 2 (1 + o(1)) for x → +∞
=
Thus
lim
x→+∞
x3 − 2x + 1
x2 + x3
2 +1
2xx−3
= e−2
Example 13.8.
log cos x
x→0 sin 2x2
lim
We have that
sin 2x2 = 2x2 + o(x4 )
x2
for x → 0
+ o(x3 ) for x → 0
cos x = 1 −
2
x2
x2
3
log 1 −
+ o(x ) = − + o(x2 ) for x → 0
2
2
Thus
lim
x→0
log cos x
1
=−
2
sin 2x
4
13-4
Example 13.9.
lim (sin x)x
2 +3x log x
x→0+
First of all we do the classic trick
2 +3x log x
lim (sin x)x
x→0+
= lim e(x
2 +3x log x) log(sin x)
x→0+
Then we note that
sin x = x + o(x2 )
for x → 0+
log sin x = log(x + o(x2 ))
= log(x(1 + o(x)))
= log x + log(1 + o(x))
for x → 0+
= log x + o(x)
Thus
lim (sin x)x
2 +3x log x
x→0+
2
= lim ex
log x+3x log2 x+o(1)
x→0+
since limx→0+ xα logβ x = 0 for any α > 0 and any real β.
Example 13.10.
lim
x→0+
log(1 + sin x)
sin 2x + x2 log x
Now we have that
sin 2x = 2x + o(x2 )
for x → 0+
log(1 + sin x) = log(1 + x + o(x2 ))
= x + o(x)
for x → 0+
sin 2x + x2 log x = 2x + o(x2 ) + x2 log x
= 2x(1 + x log x + o(x))
= 2x(1 + o(1))
for x → 0+
Thus
lim
x→0+
log(1 + sin x)
1
=
2
sin 2x + x log x
2
Example 13.11.
e2x−3 − e−3
x→0
sin x
lim
=1
13-5
We have that
e2x−3 − e−3
e2x − 1
= e−3 lim
x→0
x→0 x(1 + o(x2 ))
sin x
1 + 2x + o(x) − 1
= e−3 lim
x→0
x(1 + o(x2 ))
= 2e−3
lim
Example 13.12.
sin
√
lim
1 + x2 − 1
x
x→0
We have that
sin
lim
x→0
√
1 + x2 − 1
x
1 + 12 x2 + o(x2 ) − 1
x→0
x
1 2
2)
x
+
o(x
= lim 2
x→0
x
=0
= lim
Example 13.13.
lim
x→0
log(1 + x) + sin x + x
x + x2
2
We have that
x + x2 = x + o(x)
for x → 0
log(1 + x) = x + o(x)
for x → 0
2
sin x = x + o(x )
for x → 0
Thus
lim
x→0
Example 13.14.
log(1 + x) + sin x + x
x + x2
2
= lim
x→0
3x + o(x)
x + o(x)
1
√ 1
e− x2 + log 1 + x 5 − sin 3 x
√
√
lim
3
x→0
x−25x
2
=9
13-6
We have that
1
e− x2 = o(xα )
∀α > 0, for x → 0
√
√
√
1
1
1
log 1 + x 5 − sin 3 x = x 5 − sin 3 x + o(x 5 − sin 3 x)
√
√
√
= 5 x − 3 x + o( 5 x)
√
= 5 x(1 + o(1)) for x → 0
√
√
√
3
x − 2 5 x = −2 5 x(1 + o(1)) for x → 0
Thus
1
√ 1
√
e− x2 + log 1 + x 5 − sin 3 x
5
x(1 + o(1))
1
√
√
√
lim
=
lim
=−
3
5
5
x→0
x→0 −2 x(1 + o(1))
2
x−2 x
Example 13.15.
sin x5 3−x
lim
x→+∞
x4 2−x
We have that
lim x5 3−x = 0
x→+∞
⇒
sin x5 3−x = x5 3−x (1 + o(1))
for x → +∞
Thus
sin x5 3−x
x5 3−x (1 + o(1))
=
lim
lim
x→+∞
x→+∞
x4 2−x
x4 2−x
x
2
(1 + o(1))
= lim x
x→+∞
3
=0
since
2 x
3
= o(xα ) for any real α for x → +∞.
Example 13.16.
lim
x→+∞
cos x1
cos x2
! x2 +1
x
First of all we do the usual transformation
lim
x→+∞
cos x1
cos x2
! x2 +1
x
= lim e
x→+∞
x2 +1
x
log
1
cos x
2
cos x
13-7
then we study the exponent.
!
cos x1
1
1
1
1
log
= log 1 − 2 + o
− log 1 − 2 + o
2
3
2x
x
x
x3
cos x
1
1
1
= 2 − 2 +o
x
2x
x3
Thus
lim
x→+∞
cos
cos
1
x
2
x
! x2 +1
x
= lim e
x2 +1
x
1
+o
x2
1
x3
x→+∞
= lim e
x2 +1
+o 12
2x3
x
x→+∞
=1
Example 13.17.
√1
(2xx − 1) x − 1
√
lim
x log x
x→0+
Bearing in mind that limx→0+ x log x = 0, let us study the numerator for x → 0+ :
2xx − 1 = 2ex log x − 1
= 2(1 + x log x + o(x log x)) − 1
= 1 + 2x log x + o(x log x)
then
(1 + 2x log x + o(x log x))
√1
x
−1=e
=e
√1
x
log(1+2x log x+o(x log x))
2x log x+o(x log x)
√
x
−1
−1
√
√
2 x log x+o( x log x)
−1
√
= 1 + 2 x log x + o( x log x) − 1
√
√
= 2 x log x + o( x log x)
=e
Thus
√1
√
(2xx − 1) x − 1
√
lim
=2
x log x
x→0+