MATH 444: SOLUTIONS FOR MIDTERM 2
1(10 points):Compute
lim
x→1
√x2+ 3 −2
x−1
Note that, for x6= 1,
√x2+ 3 −2
x−1=√x2+ 3 −2
x−1·√x2+ 3 + 2
√x2+ 3 + 2 =x2−1
(x−1)(√x2+ 3 + 2)
=(x−1)(x+ 1)
(x−1)(√x2+ 3 + 2) =x+ 1
√x2+ 3 + 2,
hence
lim
x→1
√x2+ 3 −2
x−1= lim
x→1
x+ 1
√x2+ 3 + 2 =1 + 1
√12+ 3 + 2 =1
2.
2(10 points):For x∈R, define f(x) = x3+ 4x+ 1.
(a) Prove that fis strictly increasing on R.
f′(x) = 3x2+ 4 >0 for any x∈R. Thus, fis strictly increasing.
(b) Consider g=f−1(the inverse function to f). Compute g′(6).
Hint.g(6) is an integer between −2 and 2.
By trial and error, we see that f(1) = 6, hence g(6) = 1. Then
g′(6) = 1
f′(g(6)) =1
f′(1) =1
7.
3(10 points):For x∈R, define f(x) = xcos x. Prove that the function fis not
uniformly continuous on R.
Suppose, for the sake of contradiction, that fis uniformly continuous. Then there
exists δ > 0 s.t. |f(x)−f(y)|<1 whenever |x−y|< δ. Pick N∈Ns.t. π/N < δ.
Fix M > N. For 0 ≤k≤N, let xk= 2πM +kπ/N. Then x0= 2πM, and
xN= 2πM +π, hence f(xN)−f(x0) = (4M+1)π. On the other hand, |xk−xk−1|< δ
for 1 ≤k≤N, hence |f(xk)−f(xk−1)|<1, and, by the triangle inequality,
|f(xN)−f(x0)| ≤
N
X
k=1 |f(xk)−f(xk−1)|< N < M,
a contradiction.
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