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x’!a
Pick >0; then there is some ´1 such that if |x-a|
there is some ´2 such that if |x - a|
pick x with |x - a|
|f(x) +g(x) -(f(a) +g(a))| d"|f(x) -f(a)| +|g(x) -g(a)|
This gives the result. The other results are similar, but rather harder; see (Spivak 1967)
for proofs.
Note: Just as when dealing with sequences, we need to know that f/g is defined in some
neighbourhood of a. This can be shown using a very similar proof to the corresponding
result for sequences.
4.17. Proposition. Let f be continuous at a, and let g be continuous at f(a). Then g æ% f
is continuous at a
Proof. Pick >0. We must find ´ >0 such that if |x - a|
We find ´ using the given properties of f and g. Since g is continuous at f(a), there is
some ´1 > 0 such that if |y - f(a)|
is continuous at a, so there is some ´ >0 such that if |x - a|
Combining these results gives the required inequality.
4.18. Example. The function in Example 4.8 is continuous everywhere. When we first
studied it, we showed it was continuous at the  difficult point x = 0. Now we can deduce
that it is continuous everywhere else.
4.19. Example. The function f : x -’! sin3 x is continuous.
Solution. Write g(x) =sin(x) and h(x) =x3. Note that each of g and h are continuous,
and that f = g æ% h. Thus f is continuous.
x2 -a2
4.20. Example. Let f(x) =tan . Show that f is continuous at every point of its
x2 +a2
domain.
x2 - a2
Solution. Let g(x) = . Since -1
x2 + a2
for all values of x (whilst tan x is undefined when x =(2k +1)À/2 ), and the quotient is
continuous, since each term is, and since x2 + a2 =0 for any x. Thus f is continuous, since
f =tan æ%g.
1+x2
4.21. Exercise. Let f(x) =exp . Write down the domain of f, and showthat f
1-x2
is continuous at every point of its domain.
As another example of the use of the definitions, we can give a proof of Proposition 2.20
4.5. INFINITE LIMITS 37
4.22. Proposition. Let f be a continuous function at a, and let an ’!a as n ’!". Then
f(an) ’! f(a) as n ’!".
Proof. Pick >0 we must find N such that |an - f(a)|
f is continuous at a, we can find ´ such that if |x - a|
since an ’! a as n ’!"there is some N (taking ´ for our epsilon  but anything smaller,
like ´ = /2 etc would work) such that |an - a|
see that if n e"Nthen |an - f(a)|
We can consider the example f(x) =sin(1/x) with this tool.
4.23. Example. Suppose that limx’!0 sin(1/x) = l; in other words, assume, to get a con-
tradiction, that the limit exists. Let xn = 1/(Àn); then xn ’! 0 as n ’!", and so by
assumption, sin(1/xn) =sin(nÀ) =0 ’!l as n ’!". Thus, just by looking at a single se-
quence, we see that the limit (if it exists) can only be l. But instead, consider the sequence
xn =2/(4n+1)À, so again xn ’! 0as n’!". In this case, sin(1/xn) = sin((4n+1)À/2) = 1,
and we must also have l =1. Thus l does not exist.
Note: Sequences often provide a quick way of demonstrating that a function is not
continuous, while, if f is well behaved on each sequence which converges to a, then in fact
f is continuous at a. The proof is a little harder than the one we have just given, and is
left until next year.
4.24. Example. We know from Prop 4.22 together with Example 4.8 that if an ’! 0 as
n ’!", then
1
an sin ’! 0 as n ’!".
an
Prove this directly using squeezing.
Solution. The proof is essentially the same as the proof of Example 4.8. We have
1 1
0 d" an sin = |an|. sin d"|an| ’!0 as n ’!".
an an
4.5 Infinite limits
There are many more definitions and results about limits. First one that is close to the
sequence definition:
4.25. Definition. Say that limx’!" f(x) =l iff given >0, there is some K such that
whenever x>K, then |f(x) - l|
x2 +3
4.26. Example. Evaluate lim .
x’!"
3x2 +2x+1
Solution. The idea here should be quite familiar from our sequence work. We use the fact
that 1/x ’! 0 as x ’!". Thus
x2 +3 1+3/x2 1
= ’! as x ’!".
3x2 +2x+1 3+2/x +1/x2 3
38 CHAPTER 4. LIMITS AND CONTINUITY
4.27. Definition. Say that limx’!" f(x) ="iff given L>0, there is some K such that
whenever x>K, then f(x) >L.
The reason for working on proofs from the definition is to be able to check what results
of this type are trivially true without having to find it in a book. For example
4.28. Proposition. Let g(x) = 1/f(x). Then g(x) ’! 0+ as x ’!"iff f(x) ’!"as [ Pobierz caÅ‚ość w formacie PDF ]

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