MATH 205–04
Quiz #2 Solutions
1. (6 points) Prove that lim 5 − 3x = −1 by finding a satisfactory relationship between epsilon
x→2
and delta.
The statement limx→2 5 − 3x = −1 is an assertion that, for every value > 0, a value δ can be
furnished such that, if 0 < |x − 2| < δ, then |5 − 3x − (−1)| < . We may justify this assertion
by explicity determining how δ is calculated from to make this inference true.
|5 − 3x − (−1)| < |6 − 3x| < | − 3(x − 2)| < | − 3| · |x − 2| < 3|x − 2| < |x − 2| <
3
so we may declare that the choice of δ equal to
are given.
3
is sufficient to meet whatever challenge we
(
2. (4 points) Find a value for the parameter a such that the function f (x) =
5 − x if x ≤ 2
is
ax2 if x > 2
continuous everywhere.
The expressions 5 − x and ax2 are themselves both continuous on their domain, so the only
potential problem is that the piecewise function may not be continuous at the point where it
transitions between these two behaviors; thus we need simply to ensure continuity at the point
x = 2:
lim f (x) = lim+ f (x) = f (2)
x→2−
x→2
5 − 2 = a · 22 = 5 − 2
3 = 4a
3
=a
4
2−x3
4 −9x2 ,
7x
x→−∞
3. (4 points) Calculate the value of lim
or explicitly indicate that it does not exist.
We can look at the dominant term in the numerator and denominator, and conclude that for
3
−x3
x of very large magnitude, 7x2−x
4 −9x2 ≈ 7x4 , so that
2 − x3
−x3
−1
=
lim
= lim
=0
x→−∞ 7x4 − 9x2
x→+∞ 7x4
x→+∞ 7x
lim
or, alternatively, we could divide the numerator and denominator by x4 to get:
2 − x3
lim
= lim
x→+∞ 7x4 − 9x2
x→+∞
2−x3
x4
7x4 −9x2
x4
= lim
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x→+∞
2
x4
− x1
0−0
= 0.
9 =
7−0
7 − x2
Friday, February 1, 2013
MATH 205–04
Quiz #2 Solutions
4. (6 points) Let f (x) = 2x2 − 3x. Using the difference quotient, calculate the derivative f 0 (2).
f (2 + h) − f (2)
h→0
h
[2(2 + h)2 − 3(2 + h)] − (2 · 22 − 3 · 2)
= lim
h→0
h
(2 · 22 + 8h + h2 − 3 · 2 − 3h) − (2 · 22 − 3 · 2)
= lim
h→0
h
2
8h + h − 3h
= lim 8 + h − 3 = 5
= lim
h→0
h→0
h
f 0 (2) = lim
5. (2 point bonus) Use epsilon-delta methods (or an appropriate analogue thereof ) to prove, on
2 −2x
the back of this sheet, that lim+ x2x−6
= +∞.
x→3
x2 −2x
2x−6
= +∞ is an assertion that, for every value E, a value δ can be
The statement limx→3+
2 −2x
furnished such that, if 3 < x < 3+δ, then x2x−6
> E. We may justify this assertion by explicity
determining how δ is calculated from E to make this inference true.
x2 − 2x
>E
2x − 6
x2 − 2x > (2x − 6)E if x > 3
x2 − (2 + 2E)x + 6E > 0 if x > 3
p
p
2 + 2E − (2 + 2E)2 − 24E
= 1 + E − (1 + E)2 − 6E suffices
x<
2
p
So we note that δ = 1 + E − (1 + E)2 − 6E − 3 is sufficient; it’s a bit fiddly to show that this
quantity is in fact positive, though:
(E − 2)2
(E − 2)2
(1 + E − 3)2
(1 + E − 3)2
= E 2 − 4E + 4
> E 2 − 4E + 1
> E 2 + 2E + 1 − 6E
> (1 + E)2 − 6E
p
1 + E − 3 > (1 + E)2 − 6E
δ =1+E−
p
(1 + E)2 − 6E − 3 > 0
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Friday, February 1, 2013