70. If G(x) = 725 – x2, then what is lim
G(x) - G(1)
x → 1
X – 1
equal to?
(a)
2015 I
1
(b)
(0) - to
26
(d)
1
VO
5
Answers
Answer:
Using the definition of the limit, limx→a f(x), we can derive many general laws of limits, that help us to
calculate limits quickly and easily. The following rules apply to any functions f(x) and g(x) and also
apply to left and right sided limits:
Suppose that c is a constant and the limits
limx→a
f(x) and limx→a
g(x)
exist (meaning they are finite numbers). Then
1. limx→a[f(x) + g(x)] = limx→a f(x) + limx→a g(x) ;
(the limit of a sum is the sum of the limits).
2. limx→a[f(x) − g(x)] = limx→a f(x) − limx→a g(x) ;
(the limit of a difference is the difference of the limits).
3. limx→a[cf(x)] = c limx→a f(x);
(the limit of a constant times a function is the constant times the limit of the function).
4. limx→a[f(x)g(x)] = limx→a f(x) · limx→a g(x);
(The limit of a product is the product of the limits).
5. limx→a
f(x)
g(x) =
limx→a f(x)
limx→a g(x)
if limx→a g(x) 6= 0;
(the limit of a quotient is the quotient of the limits provided that the limit of the denominator is
not 0)
Example If I am given that
limx→2
f(x) = 2, limx→2
g(x) = 5, limx→2
h(x) = 0.
find the limits that exist (are a finite number):
(a) limx→2
2f(x) + h(x)
g(x)
=
limx→2(2f(x) + h(x))
limx→2 g(x)
since limx→2
g(x) 6= 0
=
2 limx→2 f(x) + limx→2 h(x)
limx→2 g(x)
=
2(2) + 0
5
=
4
5
(b) limx→2
f(x)
h(x)
(c) limx→2
f(x)h(x)
g(x)
Note 1 If limx→a g(x) = 0 and limx→a f(x) = b, where b is a finite number with b 6= 0, Then:
the values of the quotient f(x)
g(x)
can be made arbitrarily large in absolute value as x → a and thusthe limit does not exist.
If the values of f(x)
g(x)
are positive as x → a in the above situation, then limx→a
f(x)
g(x) = ∞,
If the values of f(x)
g(x)
are negative as x → a in the above situation, then limx→a
f(x)
g(x) = −∞,
If on the other hand, if limx→a g(x) = 0 = limx→a f(x), we cannot make any conclusions about
the limit.
Example Find limx→π−
cos x
x−π
.
As x approaches π from the left, cos x approaches a finite number −1.
As x approaches π from the left, x − π approaches 0.
Therefore as x approaches π from the left, the quotient cos x
x−π
approaches ∞ in absolute value.
The values of both cos x and x − π are negative as x approaches π from the left, therefore
lim
x→π−
cos x
x − π
= ∞.
More powerful laws of limits can be derived using the above laws 1-5 and our knowledge of some
basic functions. The following can be proven reasonably easily ( we are still assuming that c is a
constant and limx→a f(x) exists );
6. limx→a[f(x)]n =
-
limx→a f(x)
n
, where n is a positive integer (we see this using rule 4 repeatedly).
7. limx→a c = c, where c is a constant ( easy to prove from definition of limit and easy to see from
the graph, y = c).
8. limx→a x = a, (follows easily from the definition of limit)
9. limx→a x
n = a
n where n is a positive integer (this follows from rules 6 and 8).
10. limx→a
√n x =
√n a, where n is a positive integer and a > 0 if n is even. (proof needs a little extra
work and the binomial theorem)
11. limx→a
pn
f(x) = pn
limx→a f(x) assuming that the limx→a f(x) > 0 if n is even. (We will look at
this in more detail when we get to continuity)
Example Evaluate the following limits and justify each step:
(a) limx→3
x
3+2x
2−x+1
x−1
(b) limx→1
√3
x + 1
2(c) Determine the infinite limit (see note 1 above, say if the limit is ∞, −∞ or D.N.E.)
limx→2−
x+1
(x−2) .
Polynomial and Rational Functions
Please review the relevant parts of Lectures 3, 4 and 7 from the Algebra/Precalculus review
page. This demonstration will help you visualize some rational functions:
Direct Substitution (Evaluation) Property If f is a polynomial or a rational function and
a is in the domain of f, then limx→a f(x) = f(a). This follows easily from the rules shown above.
(Note that this is the case in part (a) of the example above)
if f(x) = P(x)
Q(x)
is a rational function where P(x) and Q(x) are polynomials with Q(a) = 0, then:
If P(a) 6= 0, we see from note 1 above that limx→a
P(x)
Q(x) = ±∞ or D.N.E. and is not equal to ±∞.
If P(a) = 0 we can cancel a factor of the polynomial P(x) with a factor of the polynomial Q(x)
and the resulting rational function may have a finite limit or an infinite limit or no limit at x = a.
The limit of the new quotient as x → a is equal to limx→a
P(x)
Q(x)
by the following observation which
we made in the last lecture:
Note 2: If h(x) = g(x) when x 6= a, then limx→a h(x) = limx→a g(x) provided the limits exist.
Example Determine if the following limits are finite, equal to ±∞ or D.N.E. and are not equal
to ±∞:
(a) limx→3
x
2−9
x−3
.
(b) limx→1−
x
2−x−6
x−1
.
(c) Which of the following is true:
1. limx→1
x
2−x−6
x−1 = +∞, 2. limx→1
x
2−x−6
x−1 = −∞, 3. limx→1
x
2−x−6
x−1 D.N.E. and is not
±∞,