Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
March 24, 2016
Quiz 14
The functions f and g are given by
f (x) = tan x
and
g(x) =
3
.
2x − π
Evaluate the following limit
lim
π −
x→( 2 )
f (x)
.
g(x)
Recall: L’Hopital’s Rule
Let f and g be differentiable on an open interval I containing a with
•
•
lim f (x) = 0
(resp. lim f (x) = ±∞ )
lim g(x) = 0
(resp. lim g(x) = ±∞ )
x→a−
x→a−
x→a
x→a
• g 0 (x) 6= 0 when x 6= a, then
f (x)
f 0 (x)
= lim 0
−
g(x)
g (x)
x→a
lim
x→a−
Since
lim
tan x = ∞
π −
and
x→( 2 )
lim
π −
x→( 2 )
3
= −∞ using
2x − π
L’Hopital’s Rule we have
lim
π −
x→( 2 )
f (x)
=
g(x)
=
lim
π −
x→( 2 )
lim
x→( π2 )−
tan x
L’Hopital’s Rule
=
lim
π −
(tan x)0
0
3
3
x→( 2 )
2x−π
2x−π
1
(2x
−
π)2
cos2 x Simplify
=
lim
−6
x→( π2 )− −6 cos2 x
(2x−π)2
since
•
•
lim (2x − π)2 = 0
x→( π
)−
2
lim (−6 cos2 x) = 0
x→( π
)−
2
• (−6 cos2 x)0 = −6 · 2 cos x sin x, and −6 · 2 cos
when x 6= π2 , using L’Hopital’s Rule we have
lim
π
x→( 2 )−
=
lim
π −
x→( 2 )
π
2
sin
(2x − π)2
((2x − π)2 )0
= lim
2
2
0
π
−6 cos x
x→( 2 )− (−6 cos x)
((2x − π)2 )0
2(2x − π)(2x − π)0
=
lim
x→( π2 )− −6 · 2 cos x(cos x)0
(−6 cos2 x)0
π
2
6= 0
=
lim
π −
2(2x − π) · 2
−6 · 2 cos x(− sin x)
lim
π −
4(2x − π)
6 · 2 cos x sin x
lim
π −
4(2x − π)
6 sin 2x
lim
π −
4 (2x − π)
6 sin 2x
x→( 2 )
Simplify
=
x→( 2 )
2 cos x sin x=sin 2x
=
x→( 2 )
=
x→( 2 )
=
4
(2x − π)
lim
6 x→( π2 )− sin 2x
since
•
•
lim (2x − π) = 0
x→( π
)−
2
lim
x→( π
)−
2
sin 2x = 0
• (sin 2x)0 = cos 2x · (2x)0 = 2 cos 2x, and 2 cos(2 π2 6= 0 when
x 6= π2 , using L’Hopital’s Rule we have
lim
π
x→( 2 )−
=
=
=
=
(2x − π)
(2x − π)0
= lim
0
sin 2x
x→( π
)− (sin 2x)
2
4
4
(2x − π)0
2
lim
=
lim
6 x→( π2 )− (sin 2x)0
6 x→( π2 )− 2 cos 2x
limx→( π2 )− 2
4
6 limx→( π2 )− 2 cos 2x
4 2
4
·
=−
6 −2
6
2
−
3
2
(since
lim
x→( π
)−
2
π
2 cos 2x = 2 cos(2 ) = −2 6= 0)
2