1. No category

Limits
Look at the graph of y = x − 1:
What is the value of y as x “approaches” 3? The graph makes it pretty clear that it just
approaches y = 3 − 1 = 2. We denote this by saying that
lim x − 1 = 2
x→3
Definition 0.1 If f (x) becomes arbitrarily close to the number L as x approaches c from
both sides, then we write
lim f (x) = L
x→c
and say “the limit of f (x) as x approaches c is L.
Note in the above that c could be ∞.
Intuition:
lim f (x) = 3
x→∞
One-sided limits
• limx→c− f (x) = L – limit from the left.
• limx→c+ f (x) = L – limit from the right.
In these cases you just approach from one side or the other.
Example: The function
(
f (x) =
−(x − 2)2 + 5 x ≤ 2
10
x>2
has the following graph:
Note the closed and open circles. In this case, they mean that the value of f at 2 is 5. Here,
lim f (x) = 10,
x→2+
lim f (x) = 5
x→2−
If both the left and right limits exist and are equal, then limx→c f (x) = L
If they both exist but are not equal, then the ordinary (or two-sided) limit does not exist.
So in the above picture, limx→2 f (x) does not exist.
Example: Suppose f (x) has the following graph:
Here, limx→−3+ f (x) = −1 and limx→−3− f (x) = −1. Why? We don’t care about the value
of f at −3, only about values near −3. So, since limx→−3+ f (x) and limx→−3− f (x) both
exist and are equal, we have limx→−3 f (x) = −1. (Even though f (−3) = 2!)
Frequently, it is the case that limx→c f (x) = f (c). In other words, to calculate the limit,
you can just plug c into the function. But, this is not always the case, as seen by the last
example.
Examples:
1. Limit of a linear function:
lim 5x − 3 = 7
x→2
To see this, plug in numbers approaching 2:
x
1.99
1.999
1.9999
f (x)
6.95
6.995
6.9995
2.01
7.05
2.001 7.005
2.0001 7.0005
You can see that as x values approach 2 from either side, the value of f approaches 7.
Note that in this case the limit is f (2).
2. Evaluate the limit:
x2 − 9
lim
x→3 x − 3
In this case, f (3) is not defined. But,
x2 − 9
(x − 3)(x + 3)
=
= x + 3 when x 6= 3
x−3
(x − 3)
Note that we are only allowed to cross out the (x − 3) term if x 6= 3. Thus for all
values of x except 3, even values very close to 3, f behaves like the linear function
x + 3. Thus
x2 − 9
lim
= lim (x + 3) = 6.
x→3 x − 3
x→3
Operations with limits
Let b and c be real numbers and n a positive integer. Then
1.
2.
3.
4.
5.
6.
7.
lim b · f (x) = b lim f (x)
x→c
x→c
lim [f (x) + g(x)] = lim f (x) + lim g(x)
hx→c
i h x→c
i
lim [f (x) · g(x)] = lim f (x) lim g(x)
x→c
x→c
x→c
x→c
limx→c f (x)
f (x)
=
, assuming lim g(x) 6= 0
x→c
x→c g(x)
limx→c g(x)
lim x = c
x→c
h
in
lim [f (x)]n = lim f (x)
x→c
x→c
h
i1/n
lim [f (x)]1/n = lim f (x)
assuming the nth root exists
lim
x→c
x→c
From these it is easy to see that if f (x) is any polynomial, then limx→c f (x) = f (c).
Note how much easier it is to calculate limits with the formula limx→c f (x) = f (c) than
with a table.
Examples:
1. Evaluate limx→2 (3x3 − 5x2 − 8x + 2).
solution:
lim (3x3 − 5x2 − 8x + 2) = 3(2)3 − 5(2)2 − 8(2) + 2 = −10
x→2
2. Evaluate limx→−1
4x−5
3−x
solution:
4x − 5
limx→−1 (4x − 5)
4(−1) − 5
−9
=
=
=
x→−1 3 − x
limx→−1 (3 − x)
3 − (−1)
4
lim
3. Evaluate limx→1
x3 −1
.
x−1
solution: In this case, I cannot just plug in 1 because that would make the denominator
0. But note that when we plug 1 into the numerator we also get 0. This means that
x − 1 must be a factor of x3 − 1.
In other words we can write x3 − 1 = (x − 1)h(x) where h(x) is some polynomial.
To find h(x) we do long division
x2 + x + 1
x−1
x3 − 1
−(x3 − x2 )
x2 − 1
− (x2 − x)
x−1
− (x − 1)
0
So, x3 − 1 = (x − 1)(x2 + x + 1). Thus,
x3 − 1
(x − 1)(x2 + x + 1)
lim
= lim
= lim (x2 + x + 1) = 3
x→1 x − 1
x→1
x→1
x−1
4. Consider the function
(
f (x) =
x2 − 3x x ≤ −1
x+5
x > −1
Does limx→−1 f (x) exist?
solution: Since the function is two different things on either side of −1, we calculate
the one-sided limits.
lim − f (x) = lim − x2 − 3x = 4
x→−1
x→−1
lim f (x) = lim + x + 5 = 4
x→−1+
x→−1
The two are equal, therefore the limit exist and is equal to 4.
5. Show that
|3x|
x→0 x
lim
does not exist.
solution: To do this, we show that the two one-sided limits are different.
First, let’s compute limx→0− . Thus we are considering negative values of x. For
negative values of x, |3x| = −3x, so
lim−
x→0
|3x|
−3x
= lim−
= lim− −3 = −3
x→0
x→0
x
x
Now, let’s compute limx→0+ . In this case we are considering positive x, so |3x| = 3x.
Thus,
lim+
x→0
|3x|
3x
= lim+
= lim− 3 = 3
x→0
x→0
x
x
The two are different, so the limit doesn’t exist.
6. Find the limits of
1
x
as x approaches ±∞.
From the graph (and our intuition) we can see that the function approches 0 as x
approaches ±∞. In other words,
1
1
= 0 = lim
x→∞ x
x→−∞ x
lim
When the limx→∞ f (x) = L, we say that f has a horizontal asymptote at y = L.
So x1 has a horizontal asymptote at y = 0.
Note that
lim+
x→0
1
= ∞,
x
lim−
x→0
1
= −∞
x
Strictly speaking, these limits don’t exist because ±∞ are not numbers. Instead,
whenever we write that a limit as x approaches c is ±∞, we are saying that the
function is “unbounded” as x → c. When this happens, we say that f (x) has a
vertical asymptote at x = c.
Continuity
Recall that if f (x) is a polynomial, then limx→c f (x) = f (c). These types of limits are easy
to calculate. This leads to the following definition.
Definition 0.2 Let c ∈ (a, b) and f (x) a function whose domain contains (a, b). then the
function f (x) is continuous at c if
lim f (x) = f (c).
x→c
Note that this implies
1. f (c) is defined,
2. the limit exists, and
3. the two are equal.
Intuition:
The graph of a continuous function is one that has no holes, jumps, or gaps. It can be
“drawn without lifting the pencil”. This is intuition only.
Example:
f (x) is not continuous at
1. x = 1 because f (1) is not defined.
2. x = 2 because limx→2 does not exist.
3. x = 4. Here the limit exists, but is not equal to f (4).
These are the three basic ways something can fail continuity.
Examples:
1. Any polynomial p(x) is continuous everywhere.
2. A rational function is one of the form
f (x) =
p(x)
q(x)
where p(x) and q(x) are polynomials. If f (x) is a rational function, it will be continuous
everywhere except where q(x) = 0 (in these places, f (x) is undefined, hence certainly
not continuous).
In general, if f (x) = p(x)
, where p and q are arbitrary, then f (x) is continuous everyq(x)
where that p and q are continuous and q is not 0.
3. Consider
x2 − 1
.
x−1
Let’s graph it. f (x) is undefined at x = 1. If x 6= 1, we get
f (x) =
(x + 1)(x − 1)
x2 − 1
=
= x + 1 – this is a line.
x−1
x−1
So the graph is
This is continuous everywhere except x = 1. So the intervals on which it is continuous
are (−∞, 1) and (1, ∞).
4. The function
1
+1
2
is continuous everywhere because x + 1 is never 0.
g(x) =
x2
5. h(x) = |x| is continuous everywhere. Why?
6. Consider
(
f (x) =
1
x
2
+1 x≤2
3−x x>2
This function is clearly continuous everywhere, except possibly at x = 2. Let’s check
x = 2.
lim f (x) =?
x→2
To calculate, we look at one-sided limits:
lim f (x) = lim+ (3 − x) = 1 (Why?)
x→2+
x→2
lim f (x) = lim−
x→2−
x→2
1
x+1
2
=2
So limx→2 f (x) does not exist, so f (x) is not continuous at x = 2.
7. Now consider
(
f (x) =
|x − 2| + 3 x < 0
x+5
x≥0
This function is clearly continuous everywhere except possibly at x = 0. Let’s check
x = 0.
lim f (x) =?
x→0
Again, we look at one-sided limits:
lim f (x) = lim+ (x + 5) = 5
x→0+
x→0
lim f (x) = lim− (|x − 2| + 3) = 5
x→0−
x→0
They are equal, so
lim f (x) = 2 = f (0)
x→0
So f is continuous at x = 0, hence continuous everywhere.
The Derivative and the Slope of Tangent Lines
When given a graph of a function, we want to be able to calculate the slope of tangent
lines.
The slope of the tangent line at a point will tell us how rapidly the function is increasing
or decreasing. We use a limit to find the slopes of tangent lines.
A secant line for a function is one that intersects it at at least two points. The idea is to
approximate the tangent line using secant lines. The slope of the tangent line will be the
limit of the slopes of the secants, if it exists.
The slope of the secant line connecting x1 to x is
f (x1 ) − f (x)
x1 − x
If we pick a point near to x and denote it x + ∆x, the slope of the secant is
f (x + ∆x) − f (x)
f (x + ∆x) − f (x)
=
(x + ∆x) − x
∆x
We should think of ∆x as being small, and note it can be positive or negative.
So ideally the slope of the tangent line is
lim
∆x→0
f (x + ∆x) − f (x)
∆x
So we make the following definition.
Definition 0.3 Let f (x) be a function. Then we define the derivative of f at x, denoted
f 0 (x), as
f (x + ∆x) − f (x)
f 0 (x) = lim
∆x→0
∆x
if that limit exists, and if it does we say that f is differentiable at x.
Notation: We will denote this a few different ways. If y = f (x), then all of the following
denote the derivative:
dy
df d
,
(f (x)), y 0 ,
.
f 0 (x),
dx dx
dx
Examples:
1. Find the derivative of f (x) = 12 x + 3 at x = 2.
solution: We calculate:
f (2 + ∆x) − f (2)
∆x→0
∆x
1
(2
+
∆x)
+ 3 − ( 12 2 + 3)
2
= lim
∆x→0
∆x
1
∆x
1
=
= 2
∆x
2
f 0 (2) =
lim
So the slope of the tangent line is m = 12 . This makes sense because y = 21 x + 3 is a
line, and the tangent line is the line itself.
2. Find the slope fo the tangent line for g(x) = 1 − x2 at x (unspecified).
solution: We have
g 0 (x) =
=
=
=
=
g(x + ∆x) − g(x)
∆x
1 − (x + ∆x)2 − (1 − x2 )
lim
∆x→0
∆x
2
1 − (x + 2x∆x + ∆x2 ) − (1 − x2 )
lim
∆x→0
∆x
−2x∆x − ∆x2
lim
∆x→0
∆x
lim (−2x − ∆x)
lim
∆x→0
∆x→0
= −2x
So the slope of the tangent line at x is −2x.
3. Find the equation of the tangent line to f (x) =
1
x
at the point (1, 1).
solution: First we have to find the slope, then we need to use it and the point (1, 1)
to find the line’s equation.
f (x + ∆x) − f (x)
∆x→0
∆x
1
−1
= lim x+∆x x
∆x→0
∆x
x+∆x
x
− x(x+∆x)
x(x+∆x)
= lim
∆x→0
∆x
f 0 (x) =
=
lim
lim
x−(x+∆x)
x(x+∆x)
∆x
1
(−∆x)
= lim
·
∆x→0 ∆x x(x + ∆x)
−1
= lim 2
∆x→0 x + x∆x
1
= − 2
x
∆x→0
So f 0 (1) = −1. Recall that y = mx + b, so here y = −x + b. We sub in the point (1, 1)
to find b:
1 = −1 + 2 =⇒ b = 2 =⇒ y = −x + 2.
Notes:
1. As you can see, calculating derivatives from the definition is difficult sometimes. We
will develop rules to avoid this.
2. Not every function is differentiable everywhere. In fact, you can’t take the derivative
at points
• of discontinuity,
• with corners,
• or with vertical tangents.
Note that saying that a function isn’t differentiable at points of discontinuity is equivalent to saying that differentiable at x =⇒ continuous at x.
We would like to avoid the limit calculation for the derivative entirely, so we state some
rules for finding the derivative directly.
1. Constant Rule:
d
(c) = 0 for any constant c
dx
2. Power Rule: If n is any real number,
d n
(x ) = nxn−1
dx
So, for example,
d
(x) = 1
dx
d 7
(x ) = 7x6
dx
3. Constant Multiple Rule:
d
d
[cf (x)] = c f (x)
dx
dx
4. Sum Rule:
d
d
d
(f (x) + g(x)) =
f (x) +
g(x)
dx
dx
dx
Examples:
1. Find the derivative of f (t) = t2/3 + 3t − 3.
solution: We use a combination of the sum rule, constant multiple rule, and power
rule:
2 2/3−1
d 2/3
(t + 3t − 3) =
t
+ 3t1−1 + 0
dt
3
2 −1/3
=
t
+3
3
2. Find the derivative of f (x) =
4
.
x3
solution: Here we use the power rule and contant multiple rule:
d
4
d
=
4x−3
3
dx x
dx
d
= 4 x−3
dx
= 4(−3)x−3−1
= −12x−4
−12
=
x4