Solutions for Worksheet #14
Math 221
1. Find the limit.
sin 4x
x→0 tan 5x
Sol. 0/0 form, so we apply L’Hospital’s rule
(a) lim
sin 4x
4 cos 4x
4
cos 4x
4
= lim
=
lim
=
.
x→0 tan 5x
x→0 5 sec2 5x
5 x→0 sec2 5x
5
lim
x
x→0 1 − cos x
Sol. 0/0 form, so we apply L’Hospital’s rule
(b) lim
1
x
= lim
x→0 sin x
x→0 1 − cos x
lim
lim−
x→0
Therefore, limx→0
cos x
(c) lim
x→0 1 − sin x
Sol.
x
1−cos x
1
1
= −∞ and lim+
= +∞.
x→0 sin x
sin x
does not exist.
cos x
cos 0
=
=1
x→0 1 − sin x
1 − sin 0
lim
xn
(d) lim x (any n)
x→∞ e
Sol. Case I. For n = 0,
1
xn
= lim x = 0
x→∞ e
x→∞ ex
lim
Case II. For n < 0,
xn
1
=
lim
= 0.
x→∞ ex
x→∞ x−n ex
Case III. For n > 0, the form is ∞/∞. So, we apply L’Hospital’s rule until we
get a deterministic form:
lim
xn
nxn−1
=
lim
x→∞ ex
x→∞
ex
lim
Notice that the denominator remains the same. The form will remain indeterminate form until we get the exponent of x to be less than equal to zero, and this
will happen after you differentiate xn bnc + 1-times; therefore, we need to apply
1
L’Hospital’s rule bnc + 1-times. And after the n applications of L’Hospital’s. At
that stage, we get back to the previous cases and we obtain,
xn
n · n − 1 · . . . (n − bnc + 1)xn−bnc−1
=
lim
= 0.
x→∞ ex
x→∞
ex
lim
Thus for any n,
xn
= 0.
x→∞ ex
lim
1
1
−
(e) lim+
x→0
x ex − 1
Sol. ∞ − ∞ form.
lim+
x→0
1
1
− x
x e −1
= lim+
x→0
ex − 1 − x
x(ex − 1)
This is 0/0 form, so we apply L’Hospital’s rule to obtain
lim+
x→0
ex − 1
ex − 1 − x
=
lim
x→0+ ex − 1 + xex
x(ex − 1)
Again 0/0 form, so by L’Hospital’s rule
ex − 1
x→0 ex − 1 + xex
1
ex
1
ex
=
lim
=
lim
=
= lim+ x
x→0+ ex (2 + x)
x→0+ x + 2
x→0 e + xex + ex
2
(f) lim+ ln(x7 − 1) − ln(x5 − 1)
lim+
x→1
Sol.
7
5
lim+ ln(x − 1) − ln(x − 1) = lim+ ln
x→1
x→1
x7 − 1
x5 − 1
x7 − 1
= ln lim+ 5
x→1 x − 1
This is 0/0 form so we apply LH rule to obtain
x7 − 1
7x6
7 2
ln lim+ 5
= ln lim+ 4 = ln lim+ x = ln(7/5) = ln 7 − ln 5.
x→1 x − 1
x→1 5x
x→1 5
Note that since ln is a continuous function on (0, ∞), it is legitimate to take the
limit inside.
√
(g) lim+ x
x
x→0
Sol. 00 form
√
lim+ x
x→0
x
= lim+ eln x
√
x
x→0
√
= lim+ e
x→0
2
x ln x
= elimx→0+
√
x ln x
In order to compute limx→0+
LH rule.
√
x ln x, we write it in the appropriate form to apply
lim+
x→0
√
ln x
x ln x = lim+ √
x→0 1/ x
Now by LH rule,
lim+
x→0
ln x
1
√ = lim
= lim −2x1/2 = 0
+
1/ x x→0 −1/2x x−3/2 x→0+
Therefore,
√
lim+ x
x
= elimx→0+
√
x ln x
= e0 = 1.
x→0
a bx
(h) lim 1 +
x→∞
x
Sol. As in the previous part,
1
a bx
lim 1 +
= elimx→∞ bx ln(1+ ax )
x→∞
x
And
1
lim bx ln 1 +
x→∞
ax
1
ln 1 + ax
= lim
x→∞
1/bx
which is 0/0 form. So we apply LH rule.
1
ln 1 + ax
−1
lim
= lim
2
x→∞
x→∞ ax (1 + 1/(ax))(−1/(bx2 ))
1/(bx)
bx2 ax
bx
b
= lim
=
lim
=
.
x→∞ ax2 (1 + ax)
x→∞ 1 + ax
a
b
Thus, the answer is e a .
2. A metal cable has radius r and is covered by insulation, so that the distance from the
center of the cable to the exterior of the insulation is R. The velocity v of an electrical
impulse in the cable is
r 2 r ln
v = −c
R
R
where c is a positive constant. Find the following limits and interpret your answers
(a) lim+ v
R→r
Sol.
lim v = lim+ −c
R→r+
r 2
R→r
R
ln
r
R
= −c ln 1 = 0.
This means the velocity v of the electrical impulse in the cable is approaching
zero as the insulation layer becomes thin.
3
(b) lim+ v
r→0
Sol.
lim+ v = lim+ −c
r 2
ln
r
R
R
ln Rr
c
c
r
2
= − 2 lim+ r ln
= − 2 lim+
R r→0
R
R r→0 1/r2
r→0
r→0
Now apply LH rule to obtain
ln Rr
c
lim v = − 2 lim+
r→0+
R r→0 1/r2
c
c
R
r3
= − 2 lim+
=
−
lim
R r→0 −2r−3 rR
R2 r→0+ −2r
c
r2
=0
= − 2 lim+
R r→0 −2
This means the velocity v of the electrical impulse in the cable is approaching
zero as the radius of the metal cable approaches zero.
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