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Limit and Continuity of Piecewise-defined Functions
When we write f(x) = |2x + 5| as f(x) =
x   52
2 x  5,

5
  (2 x  5), x   2
Page 26
, the new format is called a piecewise-
defined function because it has two (or more) pieces instead of one. And usually, when we
have a piecewise-defined function, we will be asked to find the limit at the cutoff numbers
(from one piece to another piece) and to determine whether the function is continuous at those
numbers. For example,
x
 x 2  5, x  2
Ex 1. f ( x)  
3x  1, x  2
y
x
y
h/d
x
y
 x  6,
x  2

Ex 2. f ( x)  | x  2 |,  2  x  1
 2
x 1
 x ,
y
x
1. limx2 f(x) = ___
h/d
x
y
h/d
h/d
O
y
x
y
O
2. Is f continuous at 2?
___
Note:
When we sketch the graph of a piecewise-defined function first, of course we can
tell the limit and continuity at a particular number. However, this approach often
takes longer than it really should. We also can tell the limit and continuity of a
piecewise-defined function at any number without any graphing—just do it
algebraically.
x
h/d
1. limx–2 f(x) = ___
2. Is f continuous at –2? ___
3. limx1 f(x) = ___
4. Is f continuous at 1? ___
Limit and Continuity of Piecewise-defined Functions (cont’d)
Page 27
Recall that to determine the limit at a number a, we must consider both __________________
at a. To determine whether the function is continuous at a, not only the limit at a must exist,
but also i. _____________ and ii. _____________. This is how we are going to determine the
limit and continuity of a piecewise-defined function algebraically.
x  1
| 2 x  1|,
 2
Ex 1. Given f ( x)  | x  4 |,  1  x  3 . Find the limits at x = –1 and 3 if they exist. Also determine
 2 x  1,
x3

whether f is continuous at –1 and 3 or not.
Note: Most piecewise-defined functions have the property that each piece is nice, i.e., each piece itself is continuous.
Therefore, we can use direct substitution to find the one-sided limits at a cutoff number as well as its function-value (of
course, we have to determine which piece should be used for the “plug-it-in”). Direct substitution is also feasible when
the number in question is not a cutoff number and if the piece contains that number is nice. For example, use the
function f above to find the limit at x = 1 and determine its continuity:_______________________________________.
This brings us to a classic problem when the piecewise-defined function contains a piece which is not so nice: Find the
value of k so that f(x) is continuous at x = 2.
 x 2  3x  2

f ( x)   x  2 , x  2
 k ,
x2
Limits Involving Trigonometry
Page 28
We haven’t done a lot of limit problems involving trigonometry, since we need to
recall some of the basic trig. identities and the two limits on the right. The second
one really can be derived from the first one, therefore you only need to know
limx0 (sin x)/x = 1. Let’s see why the limit is 1 (using the tabular method and a
calculator of course, however, make sure your calculator is in radian mode).
If you compare the 1st row (x) and the 2nd row (sin x),
you can see the values are really close to each other,
and the ratio between two quantities approximately
the same is approximately equal to __.
x
–.1
–.01
–.001
sin x
sin x
x

0
sin x
1
x 0 x
1. lim
cos x  1
0
x 0
x
2. lim





0.001
0.01
0.1
One thing we have to be careful is that limit of (sin x)/x is not always 1, because it also depends on the
number x is approaching. For example, the two limits on the right will not be 1.
sin x
1. lim


x

x
2
On the other hand, many cases are equal to 1, provided that it is of the form
sine of an expression
sin x
2. lim
lim

x a the same expression
x  x
and a, the number x is approaching, makes the expression = 0. This is what I mean:
sin( x   )
1. lim
1
x 
x 
sin(4 x 2  1)
2. lim1
1
x 2
4 x2  1
sin(2cos x  1)
3. lim
1
x 3
2cos x  1
cos x  1
0
x0
x
sin x
 1?
derived from lim
x0 x
How is lim
sin (expression 1)
where expression 1 and expression 2 are not the
x a
expression 2
sin(4 x 2  1)
same but if the value of a makes both expressions = 0 (for example, lim1
),
x 2
2x 1
the limit will almost never be 1. Nonetheless, most of the time, the limit does exist.
If we have lim
Limits Involving Trigonometry—cont’d
Page 29
sin (expression 1)
where expression 1 and expression 2 are not the same but Recall:
x a
expression 2
sin x
lim
if the value of a makes both expressions = 0 (e.g., limx0 (sin x)/(2x) and limx0 (sin 2x)/x), x 0 x  1
we will, with proper algebra, make expression 1 and expression 2 exactly the same. This is what we mean:
If we have lim
sin x
x  0 2x
2. lim
sin 3x
x 0 6x
4. lim
1. lim
3. lim
sin 2x
x 0
x
Another way to evaluate:
sin 2x
lim
x 0
x
sin 4x
x 0 5 x
As we can see from example 2 above, we can do either way— Make expression 2 the same as expression 1
either make x (in denominator) to become 2x or expand sin 2x if you see:
sin (expression 1)
lim
as 2 sin x cos x. On the other hand, though the expansions for
xa
expression 2
sin 3x and sin 4x (as in examples 3 and 4) exist, they are not
familiar to most of us (and they are also much harder to Make expression 1 the same as expression 2
derive). Therefore, the expansion for sine of an expression is if you see:
expression 1
lim
almost never a good way to evaluate these limits since the
xa sin (expression 2)
better way should be: Use proper algebra to make the
expression without the sine to be exactly the same as the expression with the sine (see above). Using
this method, it allows us to tackle some much harder limits:
sin(4 x 2  1)
5. lim1
x 2
2x 1
2sin( x 2  x  2)
6. lim
x 1
3x  3
Application of limx0 (sin x)/x = 1
sin x
x
 1 , what is lim
?
x 0 x
x  0 sin x
f ( x)
g ( x)
A: __ (and in general, if lim
 L, then lim
 ___)
x a g ( x)
x a f ( x)
Page 30
Q: If lim
Recall:
Make expression 2 the same as expression 1
if you see:
sin (expression 1)
lim
x a
expression 2
Knowing this, it allows us to evaluate limits with sine of an
expression in the denominator.
3x
2x  4
1. lim
2. lim
x 0 sin 2 x
x 2 sin( x 2  x  2)
Make expression 1 the same as expression 2
if you see:
expression 1
lim
x a sin (expression 2)
In previous slide, we are only saying that don’t try to expand sine of expression using trig. identities. We
are not saying that we never need to use trig. identities because sometimes we do need them, especially
for this one:
tan x
Why not just keep in mind that:
lim
tan x
x
x 0 x
and lim
lim


x 0 x
x 0 tan x
Keeping the above limit in mind, it allows us to evaluate the following limits:
tan 2 x
3x
sin x
1. lim
2. lim
3. lim
x 0 5 x
x 0 2 tan 4 x
x 0 tan x
sin 2 x
x 0 tan 3 x
4. lim
Know Your Basic Trig. Limits:
Page 31
The box on the right shows the six basic trig. limits we should know:
The 6 trig. limits we
• The limits in the 2nd and the 4th one are obtained by reciprocating the limits should keep in mind:
sin x
in the 1st and the 3rd one, respectively. And reciprocating 1 is still 1!
1.
lim
1
x 0 x
• The limit in the last one is obtained by negating the limit in the 5th one. And
negating 0 is still 0!
x
2. lim
1
x 0 sin x
On the other hand, what are the following limit:
x
x
x
tan x
1. lim
2. lim
3. lim
3. lim
1
x  0 1  cos x
x 0 1  cos x
x  0 1  cos x
x 0 x
Note: If the sign of the infinity is of no concern, we can say the 3rd limit is __ instead of ___ (After all, 1/0
x
4.
lim
1
is either + or –).
x 0 tan x
cos x  1
Examples:
5. lim
0
x

0
x
cos x  1
tan x
1. lim
2. lim
1  cos x
x 0 sin x
x  0 1  cos x
6. lim
0
x 0
x
Four more Examples
The four examples below are not previously mentioned. The first three limits have to do with trig.
powers (we can even make a generalization of these limits), while the last one requires a totally
different method (well, we kind of mention it—hint: it’s on this slide).
sin 2 x
sin 3 x
tan 3 x
tan x
1. lim 2
2. lim 2
3. lim 5
4. lim
x 0 x
x 0 x
x 0 x
x 0 2 x  sin x