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Unformatted text preview: x
− 2 cos x
lim
= lim
= lim
= − = 4 if = ±2 2.
x→0
x→0
x→0
x2
2x
2
2 59. If k = −1 then lim (k + cos x) = k + 1 = 0, so lim 60. (a) x→0 (b) 61. x→0 − cos(1/x) + 2x sin(1/x)
which does not exist (nor is it ±∞).
cos x
x
x
1
[x sin(1/x)], but lim
= lim
= 1 and lim x sin(1/x) = 0, thus
Rewrite as lim
x→0 sin x
x→0 sin x
x→0 cos x
x→0
x
[x sin(1/x)] = (1)(0) = 0
lim
x→0 sin x Apply the rule to get lim x→0 sin(1/x)
sin x
, lim
= 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between −1
x→0
(sin x)/x x→0+ x
x sin(1/x)
does not exist.
and 1 as x → +∞, so lim+
x→0
sin x
lim x→0+ CHAPTER 4 SUPPLEMENTARY EXERCISES
1. (a) f (g (x)) = x for all x in the domain of g , and g (f (x)) = x for all x in the domain of f . (b) They are reﬂections of each other through the line y = x.
(c) The domain of one is the range of the other and vice versa. (d) The equation y = f (x) can always be solved for x as a function of y . Functions with no inverses
include y = x2 , y = sin x.
(e)
(f )
2. Yes, g is continuous; this is evident from the statement about the graphs in part (b) above.
Yes, g must be diﬀerentiable (where f = 0); this can be inferred from the graphs. Note that if
f = 0 at a point then g cannot exist (inﬁnite slope). (a) For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π ; for tan x, −π/2 < x < π/2; for sec x,
0 ≤ x < π/2 or π/2 < x ≤ π . 135 Chapter 4 (b) y y y = s in 1 x 1
y = s in x
π/2 y = c os 1 x x
π 1 x y = c os x 1
y y
y = t an x
2 y = s ec 1 x y = t an 1 x y = s ec x
2
y = s ec 1 x π/2  π/2 x
π/2 x 1 2 y = s ec x (a) when the limit takes the form 0/0 or ∞/∞ (b) 3. Not necessarily; only if lim f (x) = 0. Consider g (x) = x; lim g (x) = 0. For f (x) choose
x→0 cos x
x2
cos x, x2 , and x1/2 . There are three possibilities; lim
does not exist; lim
= 0, and
x→0 x
x→0 x
1/2
x
lim
= +∞.
x→0 x2
4. In the case +∞− (−∞) the limit is +∞; in the case −∞− (+∞) the limit is −∞, because large positive
(negative) quantities are being added to large positive (negative) quantities. The cases +∞ − (+∞)
and −∞ − (−∞) are indeterminate; large numbers of opposite sign are being subtracted, and more
information about the sizes is needed. 5. (a) x = f (y ) = 8y 3 − 1; y = f −1 (x) = (b)
(c) f (x) = (x − 1)2 ; f does not have an inverse because f is not onetoone, for example
f (0) = f (2) = 1.
√
x = f (y ) = (ey )2 + 1; y = f −1 (x) = ln x − 1 = 1 ln(x − 1)
2 (d) x = f (y ) = x+1
8 1/3 = 1
(x + 1)1/3
2 y+2
x+2
; y = f −1 (x) =
y−1
x−1 ad − bc
; if ad − bc = 0 then the function represents a horizontal line, no inverse. If ad − bc = 0
(cx + d)2
ay + b
b − xd
then y = f −1 (x) =
.
then f (x) > 0 or f (x) < 0 so f is invertible. If x = f (y ) =
cy + d
xc − a 6. f (x) = 7. (a)
(b) 2 −1/3 2 −1/3
x
−y
y − y = 0. At x = 1 and y = −1, y = 2. The tangent line is
3
3
y + 1 = 2(x − 1).
Diﬀerentiating, (xy + y ) cos xy = y . With x = π/2 and y = 1 this becomes y = 0, so the equation of the
tangent line is y − 1 = 0(x − π/2) or y = 1. 8. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1 (4/5)] = 3/5,
sin[cos−1 (5/13)] = 12/13, cos[sin−1 (4/5)] = 3/5, cos[sin−1 (5/13)] = 12/13
(a) cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13)) − sin(cos−1 (4/5)) sin(sin−1 (5/13))
33
4 12 3 5
−
=
.
=
5 13 5 13
65 Supplementary Exercises 4 (b) 136 sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13))
3 12
56
45
+
=
.
=
5 13 5 13
65 9. 3 ln e2x (ex )3 + 2 exp(ln 1) = 3 ln e2x + 3 ln(ex )3 + 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2
10. Y = ln(Cekt ) = ln C + ln ekt = ln C + kt, a line with slope k and Y intercept ln C 11. (a) ex
ex
ex
= lim
= +∞
= lim
x→+∞
x→+∞
x→+∞ x2
x→+∞ 2x
x→+∞ 2
x
2
2x
2
so lim (e /x − 1) = +∞ and thus lim x (e /x − 1) = +∞
lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim
x→+∞ (b)
(c)
12. 13. x→+∞ 1
ln x
1/x
= lim 3 = ; lim
4−1
x→1 x
x→1 4x
4 x→1
lim ln x
=
x4 − 1 1
ln x
=
4−1
x→1 x
2
lim lim ax ln a = ln a x→0 y = aeax sin bx + beax cos bx and y = (a2 − b2 )eax sin bx + 2abeax cos bx, so y − 2ay + (a2 + b2 )y
= (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0.
√
√
sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y =
y + 2 sin y cos3 y = 1
−2x
x
+ 2√
= 0.
2 (1 + x2 )3/2
(1 + x2 )2
1+x 14. ln y = 2x ln 3 + 7x ln 5; 15. Find dy
dy
/y = 2 ln 3 + 7 ln 5, or
= (2 ln 3 + 7 ln 5)y
dx
dx dy
dz
= a and
= −b. From the ﬁgure
dt
dt
√
sin θ = y/z ; when x = y = 1, z = 2. So θ = sin−1 (y/z ) and
1
y dz
a
1 dy
dθ
=
−2
= −b − √ when x = y = 1.
2 /z 2
dt
z dt
z dt
2
1−y
dθ
dt 1
−2x
,y =
, hence
2
1+x
(1 + x2 )2 given x=1
y =1 z y θ x 16. (a) f (x) = −3/(x +...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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