Fall 2016, Math 409, Section 502
Solutions to some problems from the sixth assignment
Exercise 1. Use only the definition of the limit of a function to solve
the following: (iii) if a ∈ R, prove that lim x2 = a2 .
x→a
ε
Solution. Fix ε > 0 and set δ = min
, 1 . Then, if x ∈ R with
2|a| + 1
|x − a| < δ, we have
|x2 − a2 | = |x − a| · |x + a| < δ|(x − a) + 2a| 6 δ(δ + 2|a|)
ε
6 δ(1 + 2|a|) 6
(1 + 2|a|) = ε.
2|a| + 1
Exercise 2. Using only properties of limits proved so far, evaluate the limit
x2 − 2x − 3
.
x→3 x3 − 9x
lim
Solution. Note that x2 −2x−3 = (x−3)(x+1) and x3 −9x = x(x−3)(x+3)
for all x ∈ R. Hence, for all x ∈ R \ {3} we have
(x − 3)(x + 1)
(x + 1)
x2 − 2x − 3
=
=
.
x3 − 9x
x(x − 3)(x + 3)
x(x + 3)
Hence, by known properties of limits,
lim (x + 1)
x2 − 2x − 3
2
x→3
lim
=
= .
x→3
x3 − 9x
lim x( lim (x + 3))
9
x→3
x→3
Exercise 3. Let f : R → R, g : R → R be functions, and a, L, and M be
real numbers. Assume that there exists an open interval I, containing a, so
that f (x) 6= L for all x ∈ I \{a}. Then, if lim f (x) = L, and lim g(x) = M,
x→a
x→L
prove that lim g(f (x)) = M .
x→a
Solution. Since lim g(x) = M , there exists some open interval J containing
x→L
L witnessing this fact. Fix ε > 0. Then, there is some δ0 > 0 so that for all
y ∈ J \ {L} with |y − L| < δ0 we have |g(y) − M | < ε. Find a real number
δ0 > δ00 > 0 such that (L − δ00 , L + δ00 ) ⊂ (L − δ0 , L + δ0 ) ∩ J.
Since lim f (x) = L, there exists some δ > 0 such that for all x ∈ I \ {a}
x→a
with |x − a| < δ, we have |f (x) − L| < δ00 . We will show that for x ∈ I \ {a}
we have |g(f (x)) − M | < ε. Indeed, if x ∈ I \ {a} with |x − a| < δ, then
|f (x) − L| < δ00 and hence |f (x) − L| < δ0 as well as f (x) ∈ J. Furthermore,
since x ∈ I \ {a}, by assumption f (x) 6= L and therefore f (x) ∈ J \ {L}.
We conclude that |g(f (x)) − M | < ε.
Exercise 4. Let I be an open interval of R, a ∈ I and f be a functions
whose domain contains I \ {a}. Assume that for every sequence (xn )n in
I \ {a}, that converges to a, the sequence (f (xn ))n converges to some real
number, that may depend on the sequence. Prove that there exists
L ∈ R with lim f (x) = L.
x→a
Solution. By assumption, for every sequence (xn )n in I \ {a}, that converges
to a, the sequence (f (xn ))n converges to some real number. We will show
that this number is in fact the same for all such sequences. Fix an arbitrary
sequence (xn )n in I \ {a}, that converges to a and set L = limn f (xn ). We
will show that if (yn )n is any other sequence in I \ {a}, that converges to
a, then limn f (yn ) is equal to L as well. Set limn f (yn ) = L0 . Take the
sequence (wn )n , with wn = xn , if n is odd, and wn = yn , is n is even.
This is the sequence whose terms are x1 , y2 , x3 , y4 , . . .. Since both (xn )n
and (yn )n converge to a, using the definition of convergence, one can check
that (wn )n converges to a as well. By assumption, there is some M ∈ R
with limn f (wn ) = M . Note that (f (w2n ))n is a subsequence of (f (wn ))n
and therefore limn f (w2n ) = M . However, (f (w2n ))n = (f (y2n ))n and hence
(f (w2n ))n is a subsequence of (f (yn ))n and therefore limn f (w2n ) = L0 , i.e.
L0 = M . Similarly, (f (w2n−1 ))n is a subsequence of (f (wn ))n and therefore
we obtain limn f (w2n−1 ) = M . However, (f (w2n−1 ))n = (f (x2n−1 ))n and
hence (f (w2n−1 ))n is a subsequence of (f (xn ))n and therefore limn f (w2n ) =
L, i.e. L = M . In conclusion, L = L0 , i.e. limn f (yn ) = L.
We have showed that for every sequence (yn )n in I \ {a}, that converges
to a, we have limn f (yn )n = L. By the sequential characterization of the
limit of a function we obtain that lim f (x) = L.
x→a
Exercise 5. Let n ∈ N and let P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0
be a polynomial of degree n (i.e. an 6= 0). Prove that lim P (x) = +∞, if
x→+∞
an > 0, or lim P (x) = −∞, if an < 0.
x→+∞
Solution. We prove this statement by induction on n. For the base case,
let a1 , a0 be real numbers with a1 6= 0 and P (x) = a1 x + a0 . Then, since
lim x = +∞, by known properties of limits, we obtain lim a1 x + a0 =
x→+∞
x→∞
a1 ( lim x) + a0 = ±∞, depending on the sign of a1 .
x→∞
Let now n ∈ N and assume that the conclusion holds for every polynomial
of degree n and let
P (x) = an+1 xn+1 + an xn + an−1 xn−1 + · · · + a1 x + a0
be a polynomial of degree n + 1. We assume an+1 > 0, as the case an+1 < 0
is treated similarly. Define Q = an+1 xn + an xn−1 + · · · + a1 . Then Q
is a polynomial of degree n and, by the inductive assumption, we have
lim Q(x) = +∞ (since we assumed an+1 > 0). Since limx→+∞ x =
x→+∞
+∞, by know properties of limits, we obtain limx→+∞ xQ(x) = +∞ and
limx→+∞ P (x) = limx→+∞ (xQ(x) + a0 ) = +∞.