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Content:
Section 2.3: Limit Laws
Section 2.5: Continuity
Questions:
- Is there a number b such that lim x→ 2
bx^2 − 10 x + 10 + b x^2 − x − 2 exists? If so, find the value of b and the value of the limit. Solution. We are interested in what happens near x = 2. In the denominator, we get:
(2)^2 − (2) − 2 = 4 − 2 − 2 = 0,
so in order for the limit to exist, the numerator must also be 0 at x = 2. If this is not the case, then as the limit approaches 2 from one side, the denominator gets smaller and smaller and the limit approaches ∞, while from the other side, the limit approaches −∞. This means that b must be such that
b(2)^2 − 10(2) + 10 + b = 4b − 20 + 10 + b = 5b − 10 = 0
which gives that b = 2. The limit is therefore,
lim x→ 2
2 x^2 − 10 x + 12 x^2 − x − 2
= lim x→ 2
2(x^2 − 5 x + 6) (x − 2)(x + 1)
= lim x→ 2
(^2) (x − 2)(x − 3) (x − 2)(x + 1) = lim x→ 2
2(x − 3) x + 1
- Suppose that x^4 < f (x) < x^2 if |x| < 1 and x^2 < f (x) < x^4 if |x| > 1. Do the following limits exist? If so, find their values.
(a) lim x→− 1 f (x)
(b) lim x→ 1 f (x)
(c) lim x→ 0 f (x)
Solution.
(a) We have the following limits:
lim x→− 1 +
x^4 = 1; lim x→− 1 −
x^4 = 1; lim x→− 1 +
x^2 = 1; lim x→− 1 −
x^2 = 1
so by application of the Squeeze Theorem we get that
lim x→− 1 f (x) = 1.
(b) We have the following limits:
lim x→ 1 +^
x^4 = 1; lim x→ 1 −^
x^4 = 1; lim x→ 1 +^
x^2 = 1; lim x→ 1 −^
x^2 = 1
so by application of the Squeeze Theorem we get that
lim x→ 1 f (x) = 1.
(c) We have the following limits: lim x→ 0 x^4 = 0; lim x→ 0 x^2 = 0;
so by application of the Squeeze Theorem we get
lim x→ 0
f (x) = 0.
- Evaluate the following limits using limit theorems.
(a) lim x→ 10
x^2 − 100 x − 9 (b) lim x→ 10
f (x), where f (x) = x^2 for all x ̸= 10, but f (10) = 99.
(c) lim x→ 8
√ (^3) x − 2
x − 8
(d) lim x→ 8
(x − 8)(x + 2) |x − 8 | Solution.
(a) lim x→ 10
x^2 − 100 x − 9
(b) lim x→ 10 f (x) = lim x→ 10 x^2 = 100 so lim x→ 10 f (x) = 100
(c)
lim x→ 8
√ (^3) x − 2
x − 8 = lim x→ 8
x − 2) ( √^3 x − 2)( √^3 x^2 + 2 √^3 x + 4) = lim^ x→^8
√ (^3) x (^2) + 2 √ (^3) x + 4 =
(d) For finding lim x→ 8
(x − 8)(x + 2) |x − 8 |
, we use
|x − 8 | =
x − 8 if x ≥ 8 −(x − 8) if x < 8.
Let’s consider left and right hand limits.
lim x→ 8 +^
(x − 8)(x + 2) (x^ −^ 8)^
= lim x→ 8 +
(x + 2) = 10.
lim x→ 8 −^
(x − 8)(x + 2) −(x − 8)
= lim x→ 8 −^
−(x + 2) = − 10.
Since the right and left hand limits are not equal, this limit does not exist.
- Use the Squeeze Theorem to show that
lim x→ 0
p x^3 + x^2 sin
π x
Solution. Let f (x) = −
x^3 + x^2 , g(x) =
x^3 + x^2 sin(π/x), and h(x) =
x^3 + x^2. Notice the domain of f (x) and h(x) is {x ∈ R|x ≥ − 1 }, i.e. those values such that x^3 + x^2 ≥ 0. Then we know that − 1 ≤ sin(π/x) ≤ 1 and so the following holds for x ≥ − 1 :
−
p x^3 + x^2 ≤
p x^3 + x^2 sin (π/x) ≤
p x^3 + x^2 ⇒ f (x) ≤ g(x) ≤ h(x)
And since lim x→ 0 f (x) = lim x→ 0 h(x) = 0, by the Squeeze Theorem, lim x→ 0 g(x) = 0.
which means that f is continuous from the left at x = 1 for any value of b. But it also has to be continuous from the right. We find the right hand limit:
lim x→ 1 +
f (x) = lim x→ 1 +
x^2 + 2b = 1 + 2b
In order for this function to be continuous from the right, we need limx→ 1 +^ f (x) = f (1), which means that b − 1 = 1 + 2b ⇒ b = − 2. Thus, when b = − 2 , f is continuous at x = 1. □
- Give one example of a function f (x) that is continuous for all values of x except x = 3, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 3, and how you know that the discontinuity is removable. Solution. Consider, for example, the function
f (x) =
(^) x (^2) − 9 x− 3 if^ x^ ̸= 3 0 if x = 3.
Observe that, for x ̸= 3,
f (x) =
x^2 − 9 x − 3
(x + 3)(x − 3) x − 3
(x + 3)(x − 3) (x − 3) =^ x^ + 3.
and conclude:
(a) The function f is continuous for on the set(−∞, 3) ∪ (3, ∞). (b) From lim x→ 3 f (x) = lim x→ 3 (x + 3) = 6 ̸= 0 = f (3)
it follows that the function f is not continuous at x = 3.
The graph of this function:
x
y
- If f (x) = x^2 + 10 sin x, show that there is a number c such that f (c) = 1000.
Solution. Observe that the given function if continuous on R. Also, for all x ∈ R,
− 1 ≤ sin x ≤ 1 ⇒ x^2 − 10 ≤ f (x) ≤ x^2 + 10
Since
10 ≈ 10 · 3 .15 = 31. 5 , we consider an interval that contains the number 31. 5. To keep our calculations simple, take the interval [0, 40]. Now
f (0) = 0^2 + 10 sin(0) = 0 < 1000 and f (40) = 40^2 + 10 sin(40) ≤ 1600 − 10 = 1590 > 1000.
Therefore, since f is continuous on the closed interval [0, 40] and since 1000 ∈ [f (0), f (40)] = [0, 1590], by the Intermediate Value Theorem there is a number c in (0, 40) such that f (c) = 1000.
- (a) Use the Intermediate Value Theorem to show that 2 x^ = (^10) x for some x > 0.
(b) Show that the equation 2 x^ = (^10) x has no solution for x < 0. Solution. Let f (x) = 2x^ − (^10) x. Note that the domain of f is the set R{ 0 } and that on its domain, as a sum of two continuous functions, f is continuous.
(a) Since f is continuous on (0, ∞) and since lim x→ 0 +
f (x) = −∞ and lim x→∞ f (x) = ∞, but the Intermediate Value Theorem there is a ∈ (0, ∞) such that f (a) = 0. (b) For all x ∈ (−∞, 0) we have that (^10) x < 0 which implies that for all x ∈ (−∞, 0) we have that all f (x) > 0.
- Suppose f is continuous on [1, 5] and the only solutions of the equation f (x) = 6 are x = 1 and x = 4. If f (2) = 8, explain why f (3) > 6. Solution. Suppose that f (3) < 6. By the Intermediate Value Theorem applied to the continuous function f on the closed interval [2, 3], the fact that f (2) = 8 > 6 and f (3) < 6 implies that there is a number c in (2, 3) such that f (c) = 6. This contradicts the fact that the only solutions of the equation f (x) = 6 are x = 1 and x = 4. Hence, our supposition that f (3) < 6 was incorrect. It follows that f (3) ≥ 6. But f (3) ̸= 6 because the only solutions of f (x) = 6 are x = 1 and x = 4. Therefore, f (3) > 6.
