Download Tutorial 10: Limit by L`Hospital`s Rule (Solution)

2012 Summer Math Course for Direct Entry Students,
Computer Science Department, HKUST.
Tutorial 10: Limit by L’Hospital’s Rule
(Solution)
1. Find each of the following limits by using L’Hospital’s Rule:
(ln x)2
x
ax −bx
limx→0 x for
limx→0+ [x3 ln x].
(a) limx→∞
(b)
(c)
some positive real numbers a and b.
1
− x1 ].
(d) limx→0 [ ln(1+x)
3 x
) .
x2
(e) limx→∞ (1 +
(a) limx→∞
(ln x)2
x
= limx→∞
= limx→∞
= limx→∞
2 ln x
x
2
=
x
2 ln x x1
1
= limx→∞
0.
2
x
1
(b) Let y = ax .
Then, ln y = x ln a.
By differentiating both sides with respect to x,
dy
1 dy
d x
= ln a ⇒ dx
= y ln a ⇒ dx
a = (ln a)ax .
y dx
d x
b
dx
Similarly,
= (ln b)bx .
ax −bx
x
(ln a)ax −(ln b)bx
limx→0
=
1
Therefore, limx→0
=
ln a − ln b = ln ab .
(c) limx→0+ [x3 ln x] = limx→0+
= limx→0+
1
x
3
x4
−
ln x
1
x3
3
= limx→0+ − x3 = 0.
1
(d) limx→0 [ ln(1+x)
− x1 ] = limx→0
= limx→0
1
1− 1+x
1
x 1+x +ln(1+x)
= limx→0
1
1
1+(1+x) 1+x
+ln(1+x)
x−ln(1+x)
x ln(1+x)
= limx→0
x
x+(1+x) ln(1+x)
= limx→0
1
1
2+ln(1+x)
= 12 .
3 x
) ]
x2
(e) ln[limx→∞ (1 +
= limx→∞
= limx→∞
ln(1+
1
x
3
)
x2
6x2
x3 +3x
= limx→∞ [x ln(1 +
−6
1
1+ 32 x3
x
−1
x2
= limx→∞
= limx→∞
3 x
)
x2
Thus, limx→∞ (1 +
3
)]
x2
6
x
1+
3
x2
= 0.
= e0 = 1.
2. Let a be a positive real number.
loga (1+x)
= loga
x
x
limx→0 a x−1 = ln a.
(a) Prove that limx→0
(b) Prove that
e.
√
(c) Prove that limn→∞ n( n a − 1) = ln a.
(a) limx→0
loga (1+x)
x
= limx→0
1
1+x
ln a
=
= limx→0
1
ln a
=
ln e
ln a
ln(1+x)
(ln a)x
= loga e.
(b) Let y = ax − 1.
Then, x = loga (1 + y).
When x approaches 0, y approaches 0. Therefore,
x
y
limx→0 a x−1 = limy→0 log (1+y)
a
=
=
1
limy→0
1
ln e
ln a
loga (1+y)
y
=
1
loga e
(by part (a))
= ln a.
(c) Let m = n1 .
When n approaches infinity, m approaches 0. Therefore,
1
√
limn→∞ n( n a − 1) = limn→∞ a n1−1
n
= limm→0
am −1
m
= ln a
(by part (b)).
2