I
Let e>0. Then e/2>0. Since lim f(x) = L, there exists S, >0 such that, if |jc-a|<5,,
then (/(*) - L\ < e!2. Also, since limg(x) = K, there exists S 2 >0 such that, if \x-a\<8 2 , then
|g(x)- K\
< |/(jc) - L| + \g(x) - K\
[triangle inequality]
6.14
Find
As *—>3, from either the right or the left, (x~3)
2
remains positive and approaches 0. Hence,
l/(x - 3)
2 becomes larger and larger without bound and is positive. Hence, lim
-j = +co (an improper
limit).
As x-*2 from the right (that is, with x>2), x-2 approaches 0 and is positive; therefore 3/(x-2)
approaches +". However, as x-*2 from the left (that is, with x<2), x-2 approaches 0 and is
negative; therefore, 3/(jc-2) approaches -». Hence, nothing can be said about lim
people prefer to write
6.16
Find
The numerator approaches 5. The denominator approaches 0, but it is positive for x > 3 and negative for
x < 3. Hence, the quotient approaches +°° as *—» 3 from the right and approaches — °° as x—» 3 from the
left. Hence, there is no limit (neither an ordinary limit, nor +°°, nor — oo). However, as in Problem 6.15, we can
write
6.17
Evaluate lim (2x
11 - 5x" + 3x
2 + 1).
LIMITS
37
Thus,
6.18
Evaluate
approaches 2. But x approaches -oo. Therefore, the limit is -oo. (Note that the limit
will always be —oo when x—* — oo and the function is a polynomial of odd degree with positive leading
coefficient.)
6.19
Evaluate
approaches 3. At the same time, x approaches +00. So, the limit is +00. (Note that the
limit will always be +00 when x—» — oo and the function is a polynomial of even degree with positive leading
coefficient.)
6.20
Find
The numerator and denominator both approach oo. Hence, we divide numerator and denominator by x
2 , the
highest power of x in the denominator. We obtain
to indicate that the magnitude
approaches
6.15
find
But
and
and
all approach ) as x
approaches 2. At the same time x approaches +°°. Hence, the limit is +0°.
As
and
all approach 0. Hence,
and
all approach 0. Hence,
-Some
Let e>0. Then e/2>0. Since lim f(x) = L, there exists S, >0 such that, if |jc-a|<5,,
then (/(*) - L\ < e!2. Also, since limg(x) = K, there exists S 2 >0 such that, if \x-a\<8 2 , then
|g(x)- K\
< |/(jc) - L| + \g(x) - K\
[triangle inequality]
6.14
Find
As *—>3, from either the right or the left, (x~3)
2
remains positive and approaches 0. Hence,
l/(x - 3)
2 becomes larger and larger without bound and is positive. Hence, lim
-j = +co (an improper
limit).
As x-*2 from the right (that is, with x>2), x-2 approaches 0 and is positive; therefore 3/(x-2)
approaches +". However, as x-*2 from the left (that is, with x<2), x-2 approaches 0 and is
negative; therefore, 3/(jc-2) approaches -». Hence, nothing can be said about lim
people prefer to write
6.16
Find
The numerator approaches 5. The denominator approaches 0, but it is positive for x > 3 and negative for
x < 3. Hence, the quotient approaches +°° as *—» 3 from the right and approaches — °° as x—» 3 from the
left. Hence, there is no limit (neither an ordinary limit, nor +°°, nor — oo). However, as in Problem 6.15, we can
write
6.17
Evaluate lim (2x
11 - 5x" + 3x
2 + 1).
LIMITS
37
Thus,
6.18
Evaluate
approaches 2. But x approaches -oo. Therefore, the limit is -oo. (Note that the limit
will always be —oo when x—* — oo and the function is a polynomial of odd degree with positive leading
coefficient.)
6.19
Evaluate
approaches 3. At the same time, x approaches +00. So, the limit is +00. (Note that the
limit will always be +00 when x—» — oo and the function is a polynomial of even degree with positive leading
coefficient.)
6.20
Find
The numerator and denominator both approach oo. Hence, we divide numerator and denominator by x
2 , the
highest power of x in the denominator. We obtain
to indicate that the magnitude
approaches
6.15
find
But
and
and
all approach ) as x
approaches 2. At the same time x approaches +°°. Hence, the limit is +0°.
As
and
all approach 0. Hence,
and
all approach 0. Hence,
-Some