Calculus I
Notes on Limits and Continuity.
Limits of piece­wise functions: Example #1: Consider the graph of the following piece-wise function. What is happening at or around 2?
⎧ x2 , x < 2
⎪
f (x ) = ⎨ 7
,x = 2
⎪3 − x , x > 2
⎩
►
There are 3 things happening at or around x = 2 .
1) The function point (2,7),
2) The graph coming in from the right to the point (2,1), and
3) The graph coming in from the left to the point (2,4).
We can refer to the y-coordinate of the point on the function, (2,7), with just function notation, f (2) = 7 .
But, since the other 2 points are not on the function, but instead just open endpoints for pieces of the function, we can’t use function
notation for them.
This is where limits come in. For the point (2,1), we would say “the limits as x approaches 2 from the right of f(x)” = lim+ f (x) = 1 .
x →2
The + superscript on the 2 is what denotes “from the right”.
While for the point (2,4), it would be “the limits as x approaches 2 from the left of f(x)” = lim− f (x) = 4 . This time, the - superscript
x →2
on the 2 is what denotes “from the left”.
Thus all together we have:
f (2) = 7
lim+ f (x) = 1
x →2
lim f (x) = 4
x → 2−
Note that limits aren’t concerned with the actual point on the graph at x=2, but instead the point that the graph is approaching as x
approaches 2 from either the right or the left.□
Example #2: What is happening at or around x=-3 for the following function?
⎧2x + 1 , x < -3
⎪
f (x ) = ⎨ 4
, x = -3
⎪ 1
, x > -3
⎩
►
Here :
f (-3) = 4
lim+ f (x) = 1
x → -3
lim f (x) = -5
x → -3−
□
SCC:Rickman
Notes on Limits and Continuity.
Page #1 of 8
Example #3: What is happening at or around x=0 for the following function?
⎧x 3 + 2 , x ≠ 0
f (x ) = ⎨
⎩ -3 , x = 0
►
Here :
f (0) = -3
lim+ f (x) = 2
x →0
lim f (x) = 2
x → 0−
So in this case, the right-hand limit and the left-hand limit are the same value, 2. In this case, the overall limits exists and we can say
“the limit as x approaches 0 of f(x)” = lim f (x) = 2 .□
x →0
Thus,
Overall Limits
lim f (x) = L ,
x →c
if and only if,
1) lim+ f (x) = L ,
x →c
and
2) lim− f (x) = L
x →c
Example #4: What is happening at or around x=1 for the following function?
⎧ x +3 ,x ≤1
f (x ) = ⎨
2
⎩-(x − 2) , x > 1
►
Here :
f (1) = 4
lim+ f (x) = 4
x →1
lim f (x) = -1
x →1−
lim f (x) Doesn ' t exist.
x →1
□
SCC:Rickman
Notes on Limits and Continuity.
Page #2 of 8
Example #5: What is happening at or around x=-2 for the following function?
⎧ 5 + 2x , x < -2
f ( x) = ⎨
⎩-5 − 3x , x ≥ -2
►
Here :
f (-2) = 1
lim+ f (x) = 1
x → -2
lim f (x) = 1
x → -2−
lim f (x) = 1
x → -2
□
Note that in the last example, the function value and both one-sided limits are the same value. This would only happen when the
function is continuous there.
Continuity: Thus, this leads to the formal definitions of continuity.
Continuity at a Point
f(x) is continuous at x = c ,
if and only if,
lim+ f (x) = lim− f (x) = f (c) .
x →c
x →c
Continuity on an Open Interval
f(x) is continuous on the interval (a,b),
if and only if,
f(x) is continuous at all x ∈ (a, b) .
There are situations when we want to be able to talk about partial continuity.
Continuity from the Right at a Point
f(x) is continuous from the right at x = c ,
if and only if,
lim+ f (x) = f (c) .
x →c
Continuity from the Left at a Point
f(x) is continuous from the left at x = c ,
if and only if,
lim− f (x) = f (c) .
x →c
Which leads to
Continuity on a Closed Interval
f(x) is continuous on the interval [a,b],
if and only if,
1) f(x) is continuous at all x ∈ (a, b) ,
2) f(x) is continuous from the right at x = a,
and 3) f(x) is continuous from the left at x = b.
Also, for when a function is not continuous, we have these definitions.
Discontinuity at a Point
f(x) is discontinuous at x = c ,
if and only if,
f(x) is not continuous at x = c .
SCC:Rickman
Notes on Limits and Continuity.
Page #3 of 8
Removable Discontinuity at a Point
f(x) has a removable discontinuity at x = c ,
if and only if,
1) f(x) is discontinuous at x = c ,
and 2) lim+ f (x) = lim− f (x) = a real number
x →c
x →c
Removable discontinuities are associated with holes in the graph.
Non-removable Discontinuity at a Point
f(x) has a non-removable discontinuity at x = c ,
if and only if,
1) f(x) is discontinuous at x = c ,
and 2) either
a) lim+ f (x) ≠ lim− f (x) ,
x →c
x →c
b) lim+ f (x) = ∞ or -∞ ,
x →c
or c) lim− f (x) = ∞ or -∞ .
x →c
Thus, non-removable discontinuities are associated with either vertical asymptotes or jump discontinuities.
Example #6: Where is the following function continuous and where is it discontinuous. Label discontinuities as removable or nonremovable.
x 2 − 7x + 12
f (x) = 2
x + 4x − 21
►
x 2 − 7x + 12 ( x − 3)( x − 4 )
First factor. f (x) = 2
=
x + 4x − 21 ( x − 3)( x + 7 )
Thus, x ≠ 3, -7 . Now reduce. f (x) =
( x − 3)( x − 4 ) ( x − 4 )
=
. Since the reduced form would be defined at
( x − 3)( x + 7 ) ( x + 7 )
x = 3 , f(x) has a hole at
x=3. While, x=-7 would be a vertical asymptote.
Therefore, f(x) has a removable discontinuity at x=3, a non-removable discontinuity at x = -7 , and is continuous for
x ∈ ( -∞,-7 ) ∪ ( -7,3) ∪ ( 3,∞ ) .□
If we know that 2 or more functions are continuous, we can combine them to form other continuous functions as follows.
Continuity of Combined Functions
If
1) f(x) is continuous at x = c ,
2) g(x) is continuous at x = c ,
and 3) c and b are real numbers,
then the following functions are also continuous,
1) b f(x)
2) (f+g)(x)
3) (f–g)(x)
4) (fg)(x)
and 5) (f/g)(x), provided that g(c)≠0.
Also,
Continuity of Composition of Functions
If
1) g(x) is continuous at x = c ,
and 2) f(x) is continuous at x = g (c) ,
then (f◦g)(x) = f(g(x)) is continuous at x = c .
Example #7: For f(x) = x2 and g(x) = sin(x), what is also continuous by the above theorems?
►
Since f(x) = x2 and g(x) = sin(x) are continuous everywhere, the following are also continuous everywhere:
H(x)=7 x2 ; J(x) = x2 +sin(x); K(x) = sin(x) – x2 ; L(x) = x2 sin(x); M(x) = sin(x2); and N(x) = sin2(x)
sin(x)
Also, P(x) = 2 would be continuous when x ≠ 0 .□
x
SCC:Rickman
Notes on Limits and Continuity.
Page #4 of 8
Evaluating limits: A direct consequence of the definition of continuity above is that
Evaluating Limits at a Point of Continuity
If f(x) is continuous at x = c ,
then,
lim f (x) = f (c) .
x →c
Thus, if we know that a function is continuous at x = c then we can evaluate the limit by just substituting in c.
Plus note that if the function of not continuous at x = c , we probably can’t just plug in c for x, but we also can’t conclude that the
limit doesn’t exist. In these cases, we’ll just have to do same more work.
So now let’s look at limits for a few types of functions. Note you will be expected to evaluate limits for more than just these types of
functions, but this will give you some basic examples to work with.
Continuity of Polynomial Functions
Polynomials are continuous everywhere.
Thus, we can always evaluate limits of polynomials by just plugging in the value for c.
Example #8: Evaluate lim ( 3x 3 − 5x 2 + 8 ) .
x →6
►
Since 3x 3 − 5x 2 + 8 is a polynomial, it is continuous everywhere. Thus, it’s continuous at x = 6 , and we can evaluate the limit by
plugging in 6 for x.
lim ( 3x 3 − 5x 2 + 8 ) = 3(6)3 − 5(6) 2 + 8 = 476. □
x →6
Example #9: Evaluate lim− ( 5x 4 − 8x ) .
x →2
►
Since 5x 4 − 8x is a polynomial, it is continuous everywhere. Thus, it’s continuous at x = 2 , and we can evaluate the limit by
plugging in 2 for x. Note, that since it’s continuous at 2, the overall limit exists, and the left hand limit is the same as the overall limit.
lim− ( 5x 4 − 8x ) = 5(2) 4 − 8(2) = 64 □
x →2
Continuity of Rational Functions
Rational functions are continuous where they are defined.
Example #10: Evaluate lim
x → -5
x+2
.
x − 3x − 10
2
►
With rational functions you usually want to start off by factoring.
x+2
x+2
lim
= lim
x → -5 x 2 − 3x − 10
x → -5 (x − 5)(x + 2)
Since, it’s defined at x=-5, it’s continuous at x = -5 , and we can plug it -5 for x.
x+2
-5 + 2
-3
1
lim
=
=
=- □
x → -5 (x − 5)(x + 2)
(-5 − 5)(-5 + 2) -10(-3)
10
SCC:Rickman
Notes on Limits and Continuity.
Page #5 of 8
x 3 − 27
.
x → 3 x 2 − 8x + 15
Example #11: Evaluate lim
►
(x − 3) ( x 2 + 3x + 9 )
x 3 − 27
=
lim
x → 3 x 2 − 8x + 15
x →3
(x − 3)(x − 5)
In this case, the function isn’t defined when x = 3 . Therefore, we can’t just plug in 3 for x. If we did, we would get 0/0 which is
indeterminate (see below).
Instead, we need to reduce the expression. Then we will be able to plug in 3 for x.
(x − 3) ( x 2 + 3x + 9 )
x 2 + 3x + 9 (3) 2 + 3(3) + 9 9 + 9 + 9
27
lim
= lim
=
=
=x →3
x →3
(x − 3)(x − 5)
x −5
(3) − 5
-2
2
Note, that since the function reduced to an expression that was defined at x=3, the graph of the original function would have a hole in
⎛ 27 ⎞
its graph at ⎜ 3, - ⎟ .
2 ⎠
⎝
lim
□
Indeterminate and Undefined: Think about the fraction
0
8
. Of course you should already know its value, but I want to set up a pattern that I’ll use on some other
fractions.
First, let’s define the fraction to be x and clear fractions.
0
=x
8
0 = 8x
Then we can ask the question “What times 8 would give me 0?” The answer is of course 0. Thus,
Now, what about
8
0
0
8
=0.
? Let’s use the same process.
8
=x
0
8 = 0x
So, what times 0 is 8? Definitely, no real or imaginary number would work. Thus,
8
0
is truly undefined since we can’t assign it a
value. Thus, this is still undefined.
Finally, what about
0
0
?
0
=x
0
0 = 0x
What times 0 is 0? In this case, any real number would work. So we say that
0
0
is indeterminate since just based on the fraction,
we can’t determine its value. But, in the previous example, we had more information about how the expression became
limits we were able to determine the value of this
SCC:Rickman
0
0
0
0
0
0
,
. Thus, using
.
Notes on Limits and Continuity.
Page #6 of 8
Therefore, when you evaluate lim f (x) , the limit may still exist even though f(x) is not defined at x = c .
x →c
What about a situation like
Example #12: Evaluate lim+
x →4
8
0
?
x −5
.
x−4
►
This time, if we try to plug in 4, we get 0 in the denominator, but not in the numerator. Implying we are at a vertical asymptote. Thus,
x −5
.
we need to think about the graph of y =
x−4
Vertical Asymptote: x = 4
No holes
Horizontal Asymptote: y = 1
No intercepts w. H.A.
x-int = 5
5
y-int =
4
From the graph we can see that as the graph approaches the vertical asymptote from the right, it’s going down without stopping.
x −5
Therefore, lim+
= -∞ . We’ll talk more about limits that return ∞ or -∞ in a later set of notes.□
x →4 x − 4
Continuity of Trigonometric Functions
Sine and Cosine are continuous everywhere,
while tangent, cotangent, secant, and cosecant are continuous where they are defined.
Example #13: Evaluate lim [sin(x) − 4 cos(2x) + cot(3x) ] .
x→
π
6
►
⎡ ⎛ π ⎞⎤
⎡ ⎛ π ⎞⎤
⎛π⎞
⎛π⎞
⎛π⎞
⎛π⎞ 1
⎛ 1 ⎞ 0 1− 4 3
lim [sin(x) − 4 cos(2x) + cot(3x) ] = sin ⎜ ⎟ − 4 cos ⎢ 2 ⎜ ⎟ ⎥ + cot ⎢3 ⎜ ⎟ ⎥ = sin ⎜ ⎟ − 4 cos ⎜ ⎟ + cot ⎜ ⎟ = − 4 ⎜ ⎟ + =
= □
π
2
2
x→
⎝6⎠
⎝6⎠
⎝3⎠
⎝2⎠ 2
⎝ 2⎠ 1
⎣ ⎝ 6 ⎠⎦
⎣ ⎝ 6 ⎠⎦
6
⎡1 + sin(x) ⎤
.
Example #13: Evaluate lim ⎢
3π
cos(x) ⎥⎦
x→
⎣
2
Since the function is not continuous at x =
function until I can plug in
0
3π
3π
, and we would get if we tried to plug in
for x, we need to manipulate the
0
2
2
3π
for x.
2
⎛ 3π ⎞
cos ⎜ ⎟
⎡
⎤
⎡
⎤
⎡1 + sin(x) ⎤ ⎡1 − sin(x) ⎤
⎡ cos(x) ⎤
1 − sin 2 (x)
cos 2 (x)
⎝ 2 ⎠ = 0 =0 □
lim ⎢
= lim ⎢
=
⎥ = lim3π ⎢
⎥ = lim3π ⎢
⎥
⎢
⎥
⎥
π
3π
3
x→
⎣ cos(x) ⎦ ⎣1 − sin(x) ⎦ x → 2 ⎢⎣ cos(x) (1 − sin(x) ) ⎥⎦ x → 2 ⎢⎣ cos(x) (1 − sin(x) ) ⎥⎦ x → 2 ⎣1 − sin(x) ⎦ 1 − sin ⎛ 3π ⎞ 1 − (-1)
2
⎜ ⎟
⎝ 2 ⎠
SCC:Rickman
Notes on Limits and Continuity.
Page #7 of 8
Properties of Real limits: It can be shown that real limits have the following properties:
Properties of Real Limits
If
1) L, K, c and b are all real numbers,
2) lim f (x) = L,
x →c
and 3) lim g(x) = K,
x →c
Then,
1) lim [ f (x) + g(x) ] = lim [ f (x) ] + lim [ g(x) ] = L + K ,
x →c
x →c
x →c
x →c
x →c
x →c
2) lim [ f (x) − g(x) ] = lim [ f (x) ] − lim [ g(x) ] = L − K ,
3) lim b f (x) = b lim f (x) = b L ,
x →c
x →c
4) lim [ f (x) g(x) ] = lim [ f (x) ] lim [ g(x) ] = L K ,
x →c
x →c
lim [ f (x) ]
x →c
⎡ f (x) ⎤ x →c
L
and 5) lim ⎢
=
= , provided lim [ g(x) ] ≠ 0 .
x → c g(x) ⎥
x →c
[g(x)] K
⎣
⎦ lim
x →c
Limits that fail to exist: We have already seen that if lim+ f (x) ≠ lim− f (x) , then lim f (x) doesn’t exist, but there are other ways that limits don’t exist.
x →c
x →c
x →c
⎛1⎞
Example #14: Evaluate lim+ sin ⎜ ⎟ .
x →0
⎝x⎠
►
Let’s break this down. First consider what is happening to 1/x.
1
lim = ∞ (See graph to the right)
x → 0+ x
Now what happens to y = sin(x) as x → ∞ ? The graph keeps
oscillating between -1 and 1. Thus, it never settles down to a single
number.
⎛1⎞
Therefore, lim+ sin ⎜ ⎟ doesn’t exist.
x →0
⎝x⎠
⎛1⎞
We can also see this in the graph of y = sin ⎜ ⎟ .
⎝x⎠
□
There are more ways that a limit can fail to exist, but this gives you the primary cases. In general, a limit fails to exist if it doesn’t
equal a real number, but if a limit is infinite, returning ∞ or -∞, then the expected answer would be the ∞ or -∞ and not “doesn’t exist.”
Conclusion: Limits can vary greatly in its method of evaluation depending on the function and where you are evaluating the limit. These were just
a few primary examples.
Also, note that I haven’t given you the formal definition of limits yet.
SCC:Rickman
Notes on Limits and Continuity.
Page #8 of 8