Chapter 2 review, problem 77.
√
√
1 + tan x − 1 + sin x
.
Evaluate lim
x→0
x3
Solution.
√
√
√
√
√
√
1 + tan x − 1 + sin x
1 + tan x − 1 + sin x
1 + tan x + 1 + sin x
√
√
lim
= lim
x→0
x→0
x3
x3
1 + tan x + 1 + sin x
[ multiply by conjugate of numerator ]
= lim
x→0
x3
(1 + tan x) − (1 + sin x)
√
√
1 + tan x + 1 + sin x
√
√
[ let w = 1 + tan x + 1 + sin x to save space ]
(1 + tan x) − (1 + sin x)
x→0
x3 w
tan x − sin x
= lim
x→0
x3 w
sin x
− sin x cos x
= lim cos x 3 cos x
[ rewrite tan x, find common denominator ]
x→0
x w
= lim
= lim
x→0
sin x−sin x cos x
cos x
x3 w
sin x − sin x cos x
x3 w cos x
(sin x)(1 − cos x)
[ factor out sin x ]
= lim
x→0
x3 w cos x
(sin x)[2 sin2 (x/2)]
= lim
x→0
x3 w cos x
[ from half-angle formula sin2 θ = (1 − cos 2θ)/2, with θ = x/2 ]
= lim
x→0
(sin x)[2 sin2 (x/2)]
[ rewrite denominator ]
x→0 4x(x/2)2 w cos x
2 2 sin x
sin(x/2)
1
[ pull apart ]
= lim
x→0 4
x
x/2
w cos x
2 sin x
sin(x/2)
1
1
lim
lim
lim
=
x→0
x→0 w cos x
2 x→0 x
x/2
[ product rule for limits ]
= lim
1
1
(1)(1)2 lim
x→0 w cos x
2
1
1
√
= lim √
2 x→0
1 + tan x + 1 + sin x cos x
!
1
1
√
√
=
2
1 + tan 0 + 1 + sin 0 cos 0
!
1
1
√
√
=
2
1 + 0 + 1 + 0 (1)
=
=
1
.
4
[ since lim
x→0
sin x
= 1]
x
[ back-substitute w ]
[ plug in x = 0 ]
0
