MA1S11 Calculus, Tutorial Sheet 4, Solutions1
27-30 October 2015
Questions
The numbers in brackets give the numbers of marks available for the question.
1. (3) Evaluate the limits
5x2 + 1
,
x→∞ 2x2 + x − 4
lim
4x3 + x2
,
x→∞ x3 − x + 1
−3x5 + 2x3 − 2
.
x→∞ 4x5 + x3 + 4
lim
lim
4x2 + 3x − 1
x→∞ x3 + 8x − 2
lim
Solution: : The highest powers of x are the same in numerator and denominator of
the first 3 ratios. Dividing in the first ratio by x2 we get
5 + 1/x2
limx→∞ (5 + 1/x2 )
5
5x2 + 1
=
lim
==
=
2
2
2
x→∞ 2 + 1/x − 4/x
x→∞ 2x + x − 4
limx→∞ (2 + 1/x − 4/x )
2
lim
In the second ratio we divide by x3 :
4x3 + x2
4 + 1/x
limx→∞ (4 + 1/x)
lim
= lim
==
=4
x→∞ x3 − x + 1
x→∞ 1 − 1/x2 + 1/x3
limx→∞ (1 − 1/x2 + 1/x3 )
and in the third ratio we divide by x5 :
−3x5 + 2x3 − 2
−3 + 2/x2 − 2/x5
limx→∞ (−3 + 2/x2 − 2/x5 )
3
=
lim
=
=−
5
3
2
5
2
5
x→∞ 4x + x + 4
x→∞ 4 + 1/x + 4/x
limx→∞ (4 + 1/x + 4/x )
4
lim
Finally, dividing by x3 both numerator and denominator of the fourth ratio we find
that both limits again exist
4/x + 3/x2 − 1/x3
limx→∞ (4/x + 3/x2 − 1/x3 )
4x2 + 3x − 1
=
lim
=
=0
x→∞ x3 + 8x − 2
x→∞ 1 + 8/x2 − 2/x3
limx→∞ (1 + 8/x2 − 2/x3 )
lim
2. (4) For the viven values of > 0 and the limit L of the function f (x) for x → ∞,
find N > 0 such that |f (x) − L| < if x > N :
(a)
1
= 0,
x→∞ (x + 2)2
lim
= 0.01
Solution: : We need to work out what the condition |f (x) − L| < implies for
x, for the given :
1
2
(x + 2)2 < ⇔ (x + 2) > 1/
√
which further simplifies to x > 1/ − 2. With = 0.01 this evaluates to x > 8,
so for this , N = 8 will do.
1
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
(b)
3x
=
2x + 1
Solution: : We first work this out for
modulus, one has
3x
3 2x + 1 − 2 < ⇔
lim
x→∞
3
, = 0.005
2
general : Using the properties of the
− <
3
3x
− <
2x + 1 2
Now, we can add 3/2 on all sides. Then, since we are interested in the limit
where x becomes large and positive, we can multiply by the positive quantity
2x + 1 on all sides of the inequality to obtain
3
3
− (2x + 1) < 3x <
+ (2x + 1)
2
2
Subtracting 3x on all sides gives
−2x − +
3
3
< 0 < 2x + +
2
2
The second inequality is always satisfied, as it just says that x has to be larger
than some negative number. The first inequality gives the condition
2x + >
3
2
which simplifies to
1
x>
2
3
3
1
− =
− .
2
4 2
For = 0.005 this implies x > 150 − 1/2 = 299/2, and N = 299/2 will therefore
do the trick.
(c)
8x + 1
= 4, = 0.1 and = 0.01
x→∞ 2x + 5
This is completely analogous to the previous calculation: starting from
lim
− <
8x + 1
−4<
2x + 5
one obtains the condition
19
− 5/2
2
which evaluates to x > 95 − 5/2 = 185/2 for = 0.1. Hence N = 185/2 is good
enough in this case. For = 0.01 we have instead x > 950−5/2 = 1895/2. Hence
N = 1895/2 is sufficient. For arbitrary > 0 we have to choose N = 19
− 5/2
2
or larger.
x>
2
3. Determine whether the following functions are differentiable and, if so, calculate their
derivative functions for all points x in their natural domains:
f (x) = x2 ,
f (x) = 3x,
Solution: Both functions are differentiable.
(x + h)2 − x2
2hx + h2
= lim
= lim (2x + h) = 2x,
h→0
h→0
h→0
h
h
3(x + h) − 3x
3h
lim
= lim
= lim 3 = 3.
h→0
h→0 h
h→0
h
lim
3