Solutions for Calculus Quiz #2 1. (a) lim
x→0
e2x − 1 ex− 1 = lim
x→0
(ex+ 1)(ex− 1)
ex− 1 = lim
x→0ex+ 1 = e0+ 1 = 2
(b) The degree of the numerator is greater than that of the denominator, so the limit does not exist. And 2 + x2
1 − x behaves as x2
−x = −x when x → −∞.
Thus lim
x→−∞
2 + x2 1 − x = ∞.
(c) We know that −1 ≤ sin x ≤ 1.
Then
− 1
|x| ≤ sin x
x ≤ 1
|x|. Also
x→∞lim − 1
|x| = lim
x→∞
1
|x| = 0.
By Sandwich Theorem, we obtain that
x→∞lim sin x
x = 0.
(d)
x→0lim
sin x cos x
x(1 − x) = lim
x→0
sin x
x · cos x 1 − x
= (lim
x→0
sin x x )(lim
x→0
cos x 1 − x)
(since these two limits exist)
= 1 · 1
= 1.
2. (a) Let g(x) = ex, h(x) = −|x|. Then g(x) is differentiable and continuous. h(x) is continuous but it is not differentiable at x = 0. Thus f (x) = (g ◦ h)(x) is continuous for x ∈ R, and is differentiable for x ∈ R\{0}.
(b) x
x+ 1 >0 ⇒ x < −1, x > 0. Thus the domain of f (x) is D = {x ∈ R|x < −1, x >
0}, and it is continuous and differentiable in D. (f(x) is not continuous and not differentiable at x ∈ [−1, 0].)
3. Let k ∈ Z.
Consider ε = 12. If f (x) is continuous at x = k, then ∃δ > 0 such that
∀x ∈ (k − δ, k + δ) ⇒ |f (x) − k| < ε = 12. But for x ∈ (k − δ, k), f (x) = k − 1, this implies that |f (x) − k| = 1 > 12. Contradiction! Thus f (x) is not continuous at x = k.
4. Let f (x) = sin x − x.
A solution of sin x = x is a value of x for which f (x) = 0.
f(1) = sin 1 − 1 < 0, f (−1) = sin(−1) − (−1) = 1 − sin 1 > 0 since | sin x| < 1 except for x = ±π2,±3π2 , ....
And f (x) is continuous, we apply the Intermediate Value Theorem, there exists c ∈ (−1, 1) such that f (c) = 0. That is, sin c = c.
5. Let f (x) = |x|. Then f (x) is continuous. See page 174, Example 4 in the text book, f(x) is not differentiable at x = 0.
6. Let y = f (x) = −e2x2 − ex.
Then f′(x) = −2e2x− e
⇒ f′(0) = −e.
The slope of the normal line is −f′1(0) = 1e. Thus the equation of the normal line is
y− f (0) = 1
e(x − 0)
⇒ y = 1 ex.
7. f′(R) = a(k + R) − aR · 1
(k + R)2 = ak (k + R)2.
∴f′(N) = ak (k + N)2.
f(R) ≥ 0 and f′(R) ≥ 0 for R ≥ 0.
R→∞lim f(R) = lim
R→∞
aR
k+ R = lim
R→∞
a
k
R + 1 = a.
f(0) = 0, f′(0) = a k > a.
R→∞lim f′(R) = lim
R→∞
ak
(k + R)2 = 0.
We graph f (R) and f′(R) in the next page.
8. We have
( f
2g)′ = 1 2(f
g)′
= 1 2
f′g− f g′ g2 Thus
(f
2g)′(2) = 1
2(1 · 3 − (−4)(−2)
32 )
= − 5 18.
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