Calculus I, Tutorial Problem Sheet, Week 3

Limits

Q1. Consider the given graph of the function f (x). Are the following statements true or false?

1.5

0.5

0
y

-0.5

-1

-1.5

-2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x

f (x) for Q1

(a) limx→0 f (x) exists, (b) limx→0 f (x) = 0, (c) limx→0 f (x) = 1
(d) limx→1 f (x) = 1, (e) limx→1 f (x) = 0, (f) limx→a f (x) exists ∀ a ∈ (−1, 1).

Solution:
(a) true, (b) true, (c) false, (d) false, (e) false, (f) true.

Q2. Consider the given graph of the function f (x). Are the following statements true or false?

0.5

-0.5
y

-1

-1.5

-2

-1 -0.5 0 0.5 1 1.5 2 2.5 3

x

f (x) for Q2

(a) limx→2 f (x) does not exist, (b) limx→2 f (x) = 1, (c) limx→1 f (x) does not exist,
(d) limx→a f (x) exists ∀ a ∈ (−1, 1) (e) limx→a f (x) exists ∀ a ∈ (1, 3).

1
 Solution:
(a) false, (b) true, (c) true, (d) true, (e) true.

Q3. If f (x) > 0 ∀ x 6= a and limx→a f (x) = L, can we conclude that L > 0? Justify your
answer.

Solution:
No. An example is provided by f (x) = x2 with a = 0 so that L = 0 which is not positive.

Q4. Justify whether the following statement is true or false.

p
If limx→a f (x) exists then so does limx→a f (x).

Solution: False. An example is provided by f (x) = −1, with a = 0.

p
Here limx→0 f (x) exists (and is equal to −1) but f (x) is not a real function.

Q5. Calculate the following limits

3x2 cos x
(a) limx→0 (2 − x), (b) limx→−1 2x−1
, (c) limx→π/2 x sin x, (d) limx→π 1−π
.
1
Solution: (a) 2, (b) −1, (c) π/2, (d) π−1 .

Q6. Calculate the following limits

x4 −1 x3 −8 √ x−1 , 4x−x 2
(a) limx→1 x3 −1
, (b) limx→2 x4 −16
, (c) limx→1 x+3−2
(d) limx→4 √ .
2− x

Solution:
x4 −1 (x2 +1)(x+1)(x−1) (x2 +1)(x+1)
(a) limx→1 x3 −1
= limx→1 (x−1)(x2 +x+1)
= limx→1 x2 +x+1
= 4/3.
x3 −8 (x−2)(x2 +2x+4) x2 +2x+4
(b) limx→2 x4 −16
= limx→2
(x+2)(x−2)(x2 +4)
= limx→2 = 3/8. (x+2)(x2 +4)
√ √
(c) x−1
limx→1 √x+3−2 = limx→1 (√(x−1)( x+3+2)
√
x+3−2)( x+3+2)
= limx→1 (x−1)( x+3+2)
x+3−4 = 4.
√ √
2 x(4−x)(2+ x)
(d) limx→4 4x−x
√ = limx→4
2− x
√ √
(2− x)(2+ x)
= limx→4 (x(2 + x)) = 16.

Q7. Calculate the limit as x → 0 of the following

1−cos x x2 x2
(a) x2
, (b) 1−cos 2x
, (c) 1−cos 4x
.

Solution:
 2 
(a) limx→0 1−cos
x2
x
= limx→0 (1−cos x)(1+cos x)
x2 (1+cos x)
= limx→0 sin x
x
1
1+cos x = 1/2.
x2 x2 (1+cos 2x) (2x)2 (1+cos 2x)
(b) limx→0 1−cos 2x = limx→0 (1−cos 2x)(1+cos 2x) = limx→0 4 sin2 2x
= 1/2.
x2 x2 (1+cos 4x) (4x)2 (1+cos 4x)
(c) limx→0 1−cos 4x = limx→0 (1−cos 4x)(1+cos 4x) = limx→0 16 sin2 4x = 1/8.

sin(x+|x|)
Q8. Does limx→0 x
exist?
If the limit exists then find it.

2
 Solution: For x > 0, sin(x+|x|)
x = sinx2x . Hence limx→0+ sin(x+|x|)
x = limx→0+ sinx2x = limx→0+ 2 sin
2x
2x
= 2.
sin(x+|x|) sin(x+|x|)
For x < 0, x = 0. Hence limx→0− x = 0.
The left-sided and right-sided limits exist but are not equal, hence the limit does not exist.

Q9. In each case either evaluate the limit or state that no limit exists
x2 +x+12 x2 +x−12 (x2 +x−12)2 (x2 +x−12)
(a) limx→3 x−3
, (b) limx→3 x−3
, (c) limx→3 x−3
, (d) limx→3 (x−3)2
,
1−1/h2 1+1/h
(e) limh→0 1+1/h2
, (f) limh→0 1+1/h2
.

Solution:
(a) no limit exists,
x2 +x−12
(b) limx→3 x−3 = limx→3 (x+4)(x−3)
x−3 = 7.
2 2 2 (x−3)2
(c) limx→3 (x +x−12)
x−3 = limx→3 (x+4)x−3 = limx→3 (x + 4)2 (x − 3) = 0.
2
(x +x−12)
(d) (x−3)2
= (x+4)(x−3)
(x−3)2
= x+4
x−3 hence no limit exists.
1−1/h2 h2 −1
(e) limh→0 1+1/h2
= limh→0 h2 +1
= −1.
1+1/h h2 +h
(f) limh→0 1+1/h2 = limh→0 h2 +1
= 0.

Q10. Calculate the limit as x → ∞ of the following

6x+7 x2
(a) 1−2x
, (b) x2 +sin2 x
.

Solution:
6x+7 6+ x7
(a) limx→∞ 1−2x = limx→∞ 1
−2
= −3.
x
sin2 x 1
(b) First note that 0 ≤ x2
≤ x2
.
1 sin2 x
As limx→∞ x2
= 0 then by the pinching theorem limx→∞ x2
= 0.
x2 1
Thus limx→∞ x2 +sin 2x = limx→∞ 2 = 1.
1+ sin 2 x
x

Q11. Calculate the following limits

 1/x
(a) limx→∞ cos(1/x)
1+(1/x)
, (b) limx→∞ 1
x
,

(c) limx→∞ (3 + x2 ) cos(1/x), (d) limx→∞ {( x32 − cos(1/x))(1 + sin(1/x))}.

Solution:
Set u = 1/x in each case
cos(1/x) cos u
(a) limx→∞ 1+(1/x) = limu→0+ 1+u = 1.
 1/x
(b) limx→∞ x1 = limu→0+ uu = limu→0+ exp(log uu ) = limu→0+ exp(u log u) =
e0 = 1.
(c) limx→∞ (3 + x2 ) cos(1/x) = limu→0+ (3 + 2u) cos u = 3.
(d) limx→∞ ( x32 − cos(1/x))(1 + sin(1/x)) = limu→0+ (3u2 − cos u)(1 + sin u) = −1.

3
Q12. For each of the following statements, either give a proof that it is true or a counter example
to show that it is false:
(a) If g(x) > 0 ∀ x > 0 and limx→∞ (f (x) − g(x)) = 0 then limx→∞ (f (x)/g(x)) = 1.
(b) If g(x) > 0 ∀ x > 0 and limx→∞ (f (x)/g(x)) = 1 then limx→∞ (f (x) − g(x)) = 0.

Solution:
(a) False. A counter example is provided by f (x) = 1/x and g(x) = 2/x.
(b) False. A counter example is provided by f (x) = x and g(x) = x + 1.

Q13. In each case either evaluate the limit or state that no limit exists
u2 (x−2)(1−cos 3x) t−5
(a) limu→−5 5−u
, (b) limy→0 (2y −8)1/3 , (c) limx→0 2x
, (d) limt→5 t2 −25
,
√ x+2 −3x4 +x2 +1 5t3 +8t2 tan(2(x−3))
(e) limx→−2 x2 +5−3
, (f) limx→∞ −5x4 −1
, (g) limt→0 3t2 −16t4
, (h) limx→3 x−3
,
√
x+3 x2 +12−4 t2 +t−2 t3 +1
(i) limx→−3 x2 +4x+3
, (j) limx→2 x−2
, (k) limt→1 t2 −1
, (l) limt→−∞ t2 +1
.

Solution:
(a) 5/2, (b) −2, (c) 0, (d) 1/10, (e) −3/2, (f) 3/5, (g) 8/3, (h) 2,
(i) −1/2, (j) 1/2, (k) 3/2, (l) no limit exists.