M408C - Week 2 - How Do Limits Arise?
• ”In Chemistry, chemical engineering, etc, pressure at a
point is a limit concept, because one must compute average
pressure (force per unit area) over shrinking (to zero diameter)
regions of space about the point in question.”
• ”In Ecology, the notion of sustainable population is a limit
concept, because one extrapolates, often by taking a limit over
all time, thereby placing an upper bound on some population
that can be sustained by a given ecosystem.”
• In Finance, compound interest formula is given by the equation
r !k
A=P 1+
,
k
where constants P = the principal invested, r = the annual
interest rate. The variable k determines the frequency with
which interest is added to the account. As k approaches infinity,
interest is added more and more often and A approaches er P .
We say lim A = er P .
k→∞
• In Physics, the instantaneous velocity of an object is a
limit concept, which can be computed as the limit value of the
average velocities over shorter and shorter time periods that
start at the given time.
”Limit is the basic building block of Calculus!”
References: https://www.quora.com/What-are-real-life-applications-of-limits-in-calculus
References: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-19-aninteresting-limit-involving-e/
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The Velocity Problem
Problem: Suppose a ball is dropped from the Learning Tower of
Pisa (Italy) in perfect condition (no air resistance). By Galileo’s
law, the distance fallen after t seconds is given by the equation:
1 2 1
g t = 9.8 t2 = 4.9 t2.
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2
Find the velocity of the ball after 4 seconds?
s(t) =
2
Difficulty: Compute the velocity at an instant of time (t = 4).
Solution: Approximate the desired quantity by computing the
average velocity over the brief time interval from (t = 4) to (t = t∗),
where t∗ > 4 and is close to 4.
average velocity =
s(t∗) − s(4)
t∗ − 4
Time Interval
Average velocity
s(5) − s(4) 4.9(52) − 4.9(42)
4≤t≤5
=
= 44.1
5−4
1
s(4.5) − s(4) 4.9(4.52) − 4.9(42)
4 ≤ t ≤ 4.5
=
= 41.65
4.5 − 4
0.5
s(4.1) − s(4) 4.9(4.12) − 4.9(42)
4 ≤ t ≤ 4.1
=
= 39.69
4.1 − 4
0.1
s(4.01) − s(4) 4.9(4.012) − 4.9(42)
=
= 39.249
4 ≤ t ≤ 4.01
4.01 − 4
0.01
s(4.001) − s(4)
4 ≤ t ≤ 4.001
= 39.2049
4.001 − 4
s(4.0001) − s(4)
4 ≤ t ≤ 4.0001
= 39.200490
4.0001 − 4
s(4.00001) − s(4)
= 39.2000489
4 ≤ t ≤ 4.00001
4.00001 − 4
As we shorten the time period, the average velocity is becoming
closer to 39.2m/s, we say the instantaneous velocity when
t = 4 is 39.2m/s.
Change in Distance 
Instantaneous V elocity = limits of 
Change in T ime
s(t∗) − s(t)
= lim
t∗ →t
t∗ − t
(1)

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
Definition 0.1 (Limit of a function) Assume f (x) is welldefined when x is near a. Then we write
lim f (x) = L
x→a
and say “the limit of f (x), as x approaches a, equals L”, if we
can make the values of f (x) as close to L as we like by taking
x to be sufficiently close to a.
Remark 0.2 The function f (x) might or might not be defined
at x = a.
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Example 0.3 Guess
sin x
.
x→0 x
lim
1
0.995
0.99
0.985
0.98
0.975
0.97
0.965
0.96
0.955
-0.5
0
0.5
sin x
for x near 0. As x approaches
Above is the graph of y =
x
sin x
0, the function
approaches 1 (although that function is not
x
defined at 0). So we guess
sin x
=1
x→0 x
lim
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Example 0.4 Guess
lim cos(x)
x→∞
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-10
-5
0
5
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The function cos x oscillates. As x becomes bigger and bigger,
the limit x→∞
lim cos(x) doesn’t exist.
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Definition 0.5 (Left-Hand Limits) We write
lim f (x) = L,
x→a−
and say “the left-hand limit of f (x) as x approaches a, (or the
limit of f (x) as x approaches a from the left) equals L, if we can
make f (x) as close to L as we like by taking x to be sufficiently
close to a and less than a.
Definition 0.6 (Right-Hand Limits) We write
lim f (x) = L,
x→a+
and say “the right-hand limit of f (x) as x approaches a, (or
the limit of f (x) as x approaches a from the right) equals L,
if we can make f (x) as close to L as we like by taking x to be
sufficiently close to a and greater than a.
Theorem 0.7
lim f (x) = L
x→a
if and only if
lim f (x) = L and
x→a−
Remark 0.8 New notations:
• x → a− means x approaches a and x < a.
• x → a+ means x approaches a and x > a.
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lim f (x) = L.
x→a+
Example 0.9 Use the graph below to decide the left-sided, rightsided and the two-sided limits of the given function at x = -2,
2, and 3.
Reference: http://aventalearning.com/courses/CALCx-HS-A09/a/unit1/
Answer:
lim f (x) = 2;
x→−2−
lim f (x) = −3;
x→2−
lim f (x) = −1;
x→3−
lim f (x) = −3 ⇒ lim f (x) DN E
x→−2+
x→−2
lim f (x) = −∞ ⇒ lim f (x) DN E
x→2+
x→2
lim f (x) = −1 ⇒ lim f (x) = −1
x→3+
x→3
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Definition 0.10 (Infinite Limits & Vertical Asymptotes)
Let f be a function defined on both side of a, except possibly at
a itself.
• Then
lim f (x) = ∞
x→a
means the values of f(x) can be made arbitrarily large
by taking x to be sufficiently close to a, but not equal to a.
• Similarly
lim f (x) = −∞
x→a
means the values of f(x) can be made arbitrarily large
negative by taking x to be sufficiently close to a, but not
equal to a.
• The line x = a is called the vertical asymptote of the
curve y= f(x), if at least one of the following statements is
true:
lim f (x) = ∞
x→a
lim f (x) = −∞
x→a
lim f (x) = ∞
x→a−
lim f (x) = −∞
x→a−
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lim f (x) = ∞
x→a+
lim f (x) = −∞
x→a+
Example 0.11 Find
lim−
x→3
lim−
x→3
x+2
x−3
x+2
= −∞
x−3
So x = 3 is the vertical asymptote of the curve y =
10
x+2
.
x−3
Question: How to find the vertical asymptotes x = a of a given curves?
Answer: The candidates for the numbers a are the numbers
that f (x) are not defined at. Then check one of the following three
limits
lim f (x); lim− f (x); lim+ f (x)
x→a
x→a
x→a
If one of those limits are infinity (in other word, the limit doesn’t
exist), then x = a is a vertical asymptote.
Example 0.12 Find the vertical asymptotes of f (x) = tan x.
sin x
, the function tan x is not defined
cos x
π 3π 5π
when cos x = 0. So tan x is not defined at x = ± , ± , ± , · · ·
2
2
2
Also, from the graph, we can see the limits of tan x are infinity
π 3π 5π
as x approaches ± , ± , ± , · · · So the vertical asymptotes of
2
2
2
f (x) = tan x are
Answer: Since tan x =
π
x=± ,
2
x=±
3π
,
2
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x=±
5π
,···
2
Definition 0.13 (Limits at Infinity & Horizontal Asymptotes)
Let f be a function defined on some interval (a, ∞). Then
lim f (x) = L
x→∞
means the values of f(x) can be made arbitrarily close
to L by taking x to be sufficiently large.
• Let f be a function defined on some interval (−∞, a). Then
lim f (x) = L
x→−∞
means the values of f(x) can be made arbitrarily close
to L by taking x to be sufficiently large negative.
• The line y = L is called the horizontal asymptote of the
curve y= f(x), if at least one of the following statements is
true:
lim f (x) = L
lim f (x) = L
x→∞
x→−∞
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Example 0.14 Draw the graph of y = 2−x. Then find x→∞
lim 2−x.
Do the given curve have any asymptotes?
Answer:
−x
lim
2
=0
x→∞
So y = 0 is a vertical asymptote.
Remark 0.15 • To find all horizontal asymptotes, compute
the following limits x→∞
lim f (x) and lim f (x).
x→−∞
– If x→∞
lim f (x) = L, then y = L is a horizontal asymptote
of the curve y = f (x).
– If lim f (x) = M , then y = M is a horizontal asympx→−∞
tote of the curve y = f (x).
• To find all vertical asymptotes, find all values of x that f (x)
are not defined at. Assume f (x) is not defined at x = a.
Then compute the following limits
lim f (x);
x→a
lim f (x);
x→a−
lim f (x);
x→a+
If at least one of those limits does not exist, then we say
x = a is a vertical asymptote of the curve y = f (x).
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