Fall 2004 Math 151
2 Limits and Rates of Change
2.3 Limits Using Limit Laws
c
Fri, 10/Sep
2004,
Art Belmonte
• Vector limits: The vector counterparts of the sum,
difference, constant multiple, and product laws above also
hold when f and g are vector functions. (Here the product
operation involved is the dot product.)
• Two specific limits: lim c = c and lim x = a.
x→a
Summary
x→a
• THEOREM: lim f (x) = L if and only if lim f (x) = L
x→a −
x→a
and lim f (x) = L. That is, the two-sided limit exists if and
The limit laws basically say that we may interchange the limiting
process with arithmetic operations, provided the resulting limiting
expressions make sense. When they don’t, we must resort to
algebraic manipulation. In the following, a and c are real
constants. Suppose that lim f (x) and lim g(x) exist.
x→a
x→a +
only if both one-sided limits exist.
• THEOREM [monotonicity]: If f (x) ≤ g(x) for all x in an
open interval containing a (except perhaps at a itself), and if
lim f (x) = L and lim g(x) = M both exist, then L ≤ M.
x→a
x→a
• Sum Law: “Limit of sum is sum of limits.”
lim ( f (x) + g(x)) = lim f (x) + lim g(x)
x→a
x→a
• SQUEEZE THEOREM: If f (x) ≤ g(x) ≤ h(x) for all x in
an open interval containing a (except perhaps at a itself), and
if lim f (x) and lim h(x) both equal L, then we must have
x→a
x→a
• Difference Law: “Limit of difference is difference of limits.”
lim ( f (x) − g(x)) = lim f (x) − lim g(x)
x→a
x→a
x→a
x→a
lim g(x) = L as well.
x→a
x→a
f (x) ≤
&
• Product Law: “Limit of product is product of limits.”
lim ( f (x)g(x)) = lim f (x) lim g(x)
x→a
x→a
L
x→a
• Quotient Law: “Limit of quotient is quotient of limits.”
(This is provided that lim g(x) 6= 0.)
x→a
lim ( f (x)/g(x)) =
x→a
.
lim f (x)
Hand Examples
99/4
lim g(x)
x→a
x→a
• Constant Multiple Law: “Constants slide out across the
limiting operation.”
lim (c f (x)) = c lim f (x)
x→a
g(x) ≤ h(x)
↓
.
Evaluate the limit lim
x→−2
x2 + x + 1
5
and justify each step by
citing the appropriate limit laws.
x→a
Solution
• Power Law: “Limit of power is power of limit.”
n
lim f (x)n = lim f (x)
x→a
We have
x→a
Here n is a positive integer. In particular, lim x n = a n .
lim
x→a
x→−2
x2 + x + 1
5
• Root Law: “Limit of root is root of limit.”
1/n
lim f (x)1/n = lim f (x)
x→a
=
=
lim
x→−2
x2 + x + 1
(−2)2 + (−2) + 1
5
5
[power law]
[direct substitution]
= 35 = 243.
x→a
Here n is a positive integer. In particular, lim x 1/n = a 1/n .
x→a
(If n is even, it is further assumed that lim f (x) ≥ 0.)
99/8
x→a
• Direct Substitution Property (“plug-and-chug”):
For values a in the domain of a polynomial or rational
function f (x), we have lim f (x) = f (a).
Evaluate the limit lim
√
3
x→64
the appropriate limit laws.
x→a
1
√ x + 3 x . Justify each step by citing
Solution
99/16
We have
Find the limit lim
x→−3
√ lim
x +3 x
x→64
1/3
1/2
+ lim 3x
[sum law]
=
lim x
x→64
x→64
=
lim x 1/3 + 3 lim x 1/2
[constant multiple law]
√
3
=
x→64
lim x
x→64
1/3
1/3
+3
x→64
x→64
= (64) + 3(64)1/2
= 4 + 3(8) = 28.
Solution
Note that direct substitution (plug-and-chug) may not be used.
(This results in 0/0, which is meaningless.) Accordingly, factor
the rational expression involved and simplify.
1/2
lim x
x 2 − x − 12
or show that it does not exist.
x +3
[power law]
lim
x→−3
x 2 − x − 12
x +3
[direct substitution]
=
(x + 3) (x − 4)
x→−3
(x + 3)
lim (x − 4)
=
−7
=
lim
x→−3
The limit is −7.
99/15
99/21
x 2 − x + 12
or show that it does not exist.
Find the limit lim
x +3
x→−3
(h − 5)2 − 25
or show that it does not exist.
h
h→0
Find the limit lim
Solution
Solution
• As x → −3+ (x approaches −3 from the right), the
denominator approaches 0 through positive values, whereas
the numerator approaches 24. Accordingly, the rational
expression approaches ∞. In other words, the right-hand
limit is ∞.
Again, direct substitution may not be used lest 0/0 result.
Therefore, factor and simplify.
(h − 5)2 − 25
h
h→0
lim
• As x → −3− (x approaches −3 from the left), the
denominator approaches 0 through negative values, whereas
the numerator approaches 24. Accordingly, the rational
expression approaches −∞. In other words, the left-hand
limit is −∞.
99/29
√
√
2−t − 2
or show that it does not exist.
Find the limit lim
t
t→0
Stewart 99/15
Solution
1000
y
vertical
asymptote
x = −3
Direct substitution gives 0/0; no go. The technique to use is called
“rationalizing the numerator.”
√
√ !
√
√
2−t − 2
2−t + 2
(2 − t) − 2
= lim √
lim
√
√
√ t
t→0
t→0 t
2−t + 2
2−t + 2
0
=
−500
−1000
−3.5
lim
h→0
h→0
• Since the two directions of approach exhibit different
behavior, the two-sided limit does not exist. One glance at a
graph of the rational expression shows this.
500
((h − 5) − 5) ((h − 5) + 5)
h
(h − 10) h
= lim
h
h→0
= lim (h − 10) = −10
=
−3
x
−2.5
=
2
−1
lim √
√ 2−t + 2
√
−1
2
≈ −0.35
√ =−
4
2 2
t→0
The exact limit is −
purposes.
√
2
4
or approximately −0.35 for comparison
Stewart 100/54
1.5
1
0.5
99/32
Find the limit lim
x→1
exist.
2
1
− 2
x −1 x −1
0
y
or show that it does not
−0.5
−1
−1.5
−6
−5
−4
x
Solution
Direct substitution gives 10 − 20 , which is meaningless. (No, it’s
not ∞ − ∞ = 0.) Let’s combine the two fractions in the limiting
expression by finding a common denominator.
2
1
(x + 1) − 2
− 2
= lim
lim
x→1 x − 1
x→1 (x − 1) (x + 1)
x −1
(x − 1)
= lim
x→1 (x − 1) (x + 1)
1
1
=
= lim
2
x→1 x + 1
Find the limit lim
t→1
−2
MATLAB Examples
s100x46
Use the Squeeze Theorem to show that
p
π = 0.
x 3 + x 2 sin
lim
x
x→0
Solution
Since −1 ≤ sin (π/x) ≤ 1, we have
π p
p
p
≤ x3 + x2
− x 3 + x 2 ≤ x 3 + x 2 sin
x
99/42
"
−3
#
t 3 − 1 t 2 − 2t + 1
or show that it does not
,
t −1
t2 − 1
exist.
or f (x) ≤ g(x) ≤ h(x) to give the functions names. Since
lim f (x) = lim h(x) = 0, we have lim g(x) = 0 by the
Solution
Squeeze Theorem. A plot of the graphs of f , g, and h near
x = 0 is illustrative.
x→0
#
(t − 1) t 2 + t + 1 (t − 1)2
,
By factoring we have lim
t −1
t→1
(t − 1) (t + 1)
# "
3
t2 + t + 1
,t − 1 =
,0 .
= lim
t +1
2
t→1
"
x→0
x→0
%-------------------------------------------------% Stewart 100/46
%
x = linspace(-0.5, 0.5, 500);
h = sqrt(x.ˆ3 + x.ˆ2);
f = -h; g = h .* sin(pi./x);
plot(x,f,’r--’, x,g, x,h,’g-.’); grid on
xlabel(’x’); ylabel(’y’)
title(’Stewart 100/46: Squeeze Theorem in action!’)
legend(’f’, ’g’, ’h’, ’Location’, ’Northwest’)
%
100/54
echo off; diary off
Find the limit
lim
x→−4−
|x + 4|
or show that it does not exist.
x +4
Stewart 100/46: Squeeze Theorem in action!
0.8
0.6
0.4
Solution
f
g
h
0.2
lim
x→−4−
A graph of
|x + 4|
=
x +4
lim
x→−4−
− (x + 4)
=
x +4
y
Since x approaches −4 from the left (x < −4), we have
x + 4 < 0. Thus |x + 4| = − (x + 4). Therefore,
0
−0.2
−0.4
lim (−1) = −1.
x→−4−
−0.6
−0.8
−0.5
|x + 4|
makes this left-hand limit clear.
x +4
3
0
x
0.5
A power tool: the limit command
s100x68

 x
Let h(x) =
x2

8−x
if x < 0,
if 0 < x ≤ 2,
if x > 2.
Here are transcripts of the hand and MATLAB examples above
dispatched via the limit command.
%-------------------------------------------------% Stewart 99/4
%
syms x
val = limit((xˆ2 + x + 1)ˆ5, x, 1)
(a) Evaluate each limit or state that it does not exist.
(ii) lim h(x)
(iii) lim h(x)
(i) lim h(x)
x→0+
x→0
x→1
val =
(iv) lim h(x)
(v) lim h(x)
x→2−
x→2+
(vi) lim h(x)
x→2
243
(b) Sketch the graph of h.
%
echo off; diary off
%-------------------------------------------------% Stewart 99/8
%
syms x
val = limit(xˆ(1/3) + 3*sqrt(x), x, 64)
Solution
(i) lim h(x) = 0
(ii) lim h(x) = 0
x→0+
x→0
(iv) lim h(x) = 4
(v) lim h(x) = 6
x→2−
x→2+
(iii) lim h(x) = 1
val =
x→1
64ˆ(1/3)+24
(vi) lim h(x) DNE
x→2
simplify(val)
ans =
%-------------------------------------------------% Stewart 100/68
%
% MAIN ACTION
x = [linspace(-3, -0.15, 300) ...
linspace(0.15, 1.96, 200)...
linspace(2.15, 3, 100)];
h = (x<0) .* x + ...
(0<x & x<=2) .* x.ˆ2 + ...
(x>2) .* (8-x);
plot(x,h,’.’); grid on
axis equal; axis([-3 3 -3 6])
% EXTRA FLUFF
hold on
plot(0,0,’o’, ’MarkerSize’, 8)
plot(2,6,’o’, ’MarkerSize’, 8)
plot(2,4,’o’, ’MarkerSize’, 8, ...
’MarkerFaceColor’, ’b’)
plot([-3 3], [0 0], ’r--’)
plot([0 0], [-3 6], ’r--’)
title(’Stewart 100/68’)
set(gca,’XTick’, -3:3)
set(gca,’YTick’, -3:6)
%
28
%
echo off; diary off
%-------------------------------------------------% Stewart 99/15 (symbolically)
%
syms x
f = (xˆ2 - x + 12) / (x + 3); pretty(f)
2
x - x + 12
----------x + 3
left limit = limit(f, x, -3, ’left’)
left limit =
-Inf
right limit = limit(f, x, -3, ’right’)
echo off; diary off
right limit =
Inf
Stewart 100/68
6
5
two sided limit = limit(f, x, -3) % DNE
h(x)
two sided limit =
4
NaN
3
2
% NaN means "Not a Number." This signifies that the
% two-sided limit does not exist.
%
echo off; diary off
%-------------------------------------------------% Stewart 99/16
%
syms x
f = (xˆ2 - x - 12) / (x + 3); pretty(f)
1
0
x
−1
−2
−3
−3 −2 −1
0
1
2
3
4
2
x - x - 12
----------x + 3
[ 3
[t - 1
[-----[ 2
[t - 1
val = limit(f, x, -3)
2
]
t - 2 t + 1]
------------]
t - 1
]
]
val = limit(r, t, 1)
val =
val =
-7
[ 3/2,
%
echo off; diary off
%-------------------------------------------------% Stewart 99/21
%
syms h
f = ((h-5)ˆ2 - 25) / h; pretty(f)
0]
%
echo off; diary off
%-------------------------------------------------% Stewart 100/46 (symbolically)
%
syms x
g = sqrt(xˆ3 + xˆ2) * sin(pi/x);
pretty(g)
2
(h - 5) - 25
------------h
3
(x
val = limit(f, h, 0)
2 1/2
pi
+ x )
sin(----)
x
val = limit(g, x, 0)
val =
val =
-10
0
%
echo off; diary off
%-------------------------------------------------% Stewart 99/29
%
syms t
f = (sqrt(2-t) - sqrt(2)) / t; pretty(f)
%
echo off; diary off
%-------------------------------------------------% Stewart 100/54 (symbolically)
%
syms x
val = limit(abs(x+4) / (x+4), x, -4, ’left’)
1/2
1/2
(2 - t)
- 2
----------------t
val = limit(f, t, 0); pretty(val)
val =
-1
%
1/2
- 1/4 2
val floated = eval(val) % or use double(val)
val floated =
-0.3536
%
echo off; diary off
%-------------------------------------------------% Stewart 99/32
%
syms x
f = 1/(x-1) - 2/(xˆ2-1); pretty(f)
echo off; diary off
1
2
----- - -----x - 1
2
x - 1
val = limit(f, x, 1)
val =
1/2
%
echo off; diary off
%-------------------------------------------------% Stewart 99/42
%
syms t
r = [(tˆ3 - 1) / (tˆ2 - 1),
(tˆ2 - 2*t + 1) / (t - 1)];
pretty(r)
%
%
%
%
5