Math 125
Carter
Test 1 Fall 2012
General Instructions: Write your name on only the outside of your blue book. Put your
test paper inside your blue book as you leave. Solve each of the following problems (105 total
points). Point values are indicated on the problems. Remember to use your turn signals
when changing lanes.
1. (5 points) What does the expression
lim f (x) = L
x→c
mean? State the definition of a limit.
Solution: We say that limx→c f (x) = L if and only if for every > 0 there is a δ > 0
such that
|f (x) − L| < provided
|x − c| < δ.
In more symbols, limx→c f (x) = L if and only if ∀ > 0, ∃δ > 0 such that
|f (x) − L| < provided
|x − c| < δ.
See also
http://www.youtube.com/watch?v=wSTJLUAu5ZI
2. (10 points) Give a proof that
lim+
h→0
sin (h)
= 1.
h
http://www.youtube.com/watch?v=o6S6RbfhRTU
See also p. 83 of the text.
3. (5 points) Use the fact that limh→0
algebra to prove that
sin (h)
h
= 1, the Pythagorean identity, and elementary
cos (h) − 1
= 0.
h→0
h
lim
1
Solution.
cos (h) − 1
(cos (h) − 1) (cos (h) + 1)
= lim
·
h→0
h→0
h
h
(cos (h) + 1)
2
(cos (h) − 1)
= lim
h→0 h(cos (h) + 1)
− sin2 (h)
= lim
h→0 h(cos (h) + 1)
−1
sin (h) sin (h)
= lim
·
·
h→0 (cos (h) + 1)
(h)
1
−1
=
·1·0
2
= 0.
lim
4. (10 points) Sketch the graph of
f (x) = 3 cos (2x)
over two full periods.
3
2
1
-3
-2
1
-1
2
3
-1
-2
-3
5. (10 points) Sketch the graph of the linear fractional transformation
f (x) =
2x − 3
.
x+4
8
6
4
2
-20
10
-10
-2
-4
2
20
6. (5 points) Use the properties of exponentials and logarithms to simplify the expression
ln (ex
2 +y 2
).
Solution.
2 +y 2
ln (ex
) = x2 + y 2 .
7. Use the rules for computing the limit for each of the following problems (10 points
each).
(a)
x2 + x − 6
x→2 x2 − 3x + 2
lim
Solution.
(x − 2)(x + 3)
x2 + x − 6
=
lim
x→2 (x − 2)(x − 1)
x→2 x2 − 3x + 2
(x + 3)
= lim
x→2 (x − 1)
5
=
1
lim
(b)
lim
x→0
Solution.
sin (2x)
sin (3x)
sin (2x)
2 sin (2x)
3x
2
= lim ·
·
= .
x→0 sin (3x)
x→0 3
2x
sin (3x)
3
lim
(c)
√
lim
x→16
Solution.
x−4
x − 16
√
√
x−4
x−4
1
√
= .
lim
= lim √
x→16 x − 16
x→16 ( x − 4)( x + 4)
8
8. (10 points) In the figure below, a rectangle is inscribed under the parabola y = 12 − x2
with its base along the x-axis and its two upper vertices on the parabola. Express the
area of the rectangle as a function of x which is the horizontal coordinate of the lower
right vertex.
3
y
y = 12 - x 2
x
x
Solution.
A = bh = 2x · (12 − x2 ).
9. (10 points) Use the definition of the derivative, f 0 (x) = limh→0
(x)
limz→x f (z)−f
), to compute the derivative of the function
z−x
f (x+h)−f (x)
h
f (x) = 2x2 + 3x − 1.
Solution.
f (x) = 2x2 + 3x − 1;
f (x + h) = 2(x + h)2 + 3(x + h) − 1
= 2x2 + 4xh + 2h2 + 3x + 3h − 1;
f (x) = 2x2
+ 3x
− 1;
2
f (x + h) − f (x) =
4xh + 2h
+ 3h;
f (x + h) − f (x)
= 4x + 2h + 3;
h
f (x + h) − f (x)
lim
= 4x + 3;
h→0
h
f 0 (x) = 4x + 3.
4
(or f 0 (x) =
10. (10 points) Use the definition of the derivative (as given above) to compute the equation
of the line tangent to the curve
1
y=
x
1
at the point (2, 2 ).
f (x) =
f (x + h) =
f (x + h) − f (x) =
=
=
f (x + h) − f (x)
=
h
f (x + h) − f (x)
=
lim
h→0
h
f 0 (2) =
1
.
x
1
x + h.
1
1
−
x+h x
x − (x + h)
x(x + h)
−h
.
x(x + h)
−1
x(x + h)
−1
.
x2
−1
.
4
The equation of the line tangent to the curve at (2, 12 ) is
1
−1
(y − ) =
(x − 2).
2
4
5