Name: SOLUTION
Math 1A Quiz #4
Section:
Each problem is worth 3 points. Show all your work to receive full credit.
1. Find the limit
lim
x→0
sin 4x
.
sin 6x
We know that limx→0 sin x/x = 1, and so we manipulate this limit to obtain similar expressions:
lim
x→0
sin 4x
sin 6x
=
4
(6x) sin 4x
· lim
6 x→0 (4x) sin 6x
=
2 limx→0
·
3 limx→0
sin 4x
4x
sin 6x
6x
sin u
limu→0 u
limv→0 sinv v
2
·
using the substitutions u = 4x, v = 6x
3
2 1
=
·
3 1
2
=
.
3
(Note that we’re brushing over the use of certain limit laws. Note also that this would be a great instance in
which to use l’Hopital’s rule, but we don’t know it yet!)
=
2. If h(x) =
p
4 + 3f (x) where f (1) = 7 and f 0 (1) = 4, find h0 (1).
We apply the chain rule to find h0 (x):
1
1
d
1
(4 + 3f (x)) 2 = · (4 + 3f (x))− 2 · (3 · f 0 (x)).
dx
2
Then, we plug in x = 1 and compute:
h0 (x) =
h0 (1) =
1
1
1
6
1
1 1
· (4 + 3 · f (1))− 2 · (3 · f 0 (1)) = · (4 + 3 · 7)− 2 · (3 · 4) = · · 12 = .
2
2
2 5
5
3. Find an equation of the tangent line to the curve y = ex cos x at (x, y) = (0, 1).
We first find the slope of the curve at the indicated point, which we do by computing the derivative of the
curve:
y 0 = (ex )0 · (cos x) + (ex ) · (cos x)0 = ex · cos x + ex · (− sin x) = ex · (cos x − sin x)
y 0 (0) = e0 · (cos 0 − sin 0) = 1 · (1 − 0) = 1.
Thus the equation of the tangent line is y − 1 = 1 · (x − 0), or more simply y = x + 1.