Calculus I, Fall 2014 (http://www.math.nthu.edu.tw/~wangwc/)

Solutions in Quiz01

1. ∃ϵ > 0, such that ∀δ > 0, ∃x_{0} ∈ (c − δ, c) ∪ (c, c + δ) and |f(x_{0}) − L| ≥ ϵ

2. lim_{θ→0} \frac{sin(sinθ)}{tan2θ} = lim_{θ→0} \frac{sin(sinθ)}{sin2θ} · cos2θ = lim_{θ→0} \frac{sin(sinθ)}{sinθ} · \frac{cos2θ}{2cosθ} = \frac{1}{2}

3. (i) Intermediate Value Theorem:

If f is a continuous function on a closed interval [a, b], and if y_{0} is any value between f(a) and f(b), then y_{0} = f(c) for some c in [a, b].

(ii) Let g(x) = x − 1 − cosx, then g is continuous on R. Since

g(1) = −cos1 < 0
g(2) = 1 − cos2 > 0

By Intermediate Value Theorem, there exist x_{0} ∈ [1, 2] such that g(x_{0}) = 0, i.e. x_{0} − 1 = cosx_{0}. Hence c = 1.

4. Since f(x) and g(x) are continuous at x = c, we have ∀ϵ > 0, ∃δ_{1} > 0, δ_{2} > 0, such that for all x,

\{ \begin{matrix}|x − c| < δ_{1} ⟹ |f(x) − f(c)| < \frac{ϵ}{5} \\ |x − c| < δ_{2} ⟹ |g(x) − g(c)| < \frac{ϵ}{5} \end{matrix}

Taking δ = min(δ_{1}, δ_{2}) > 0, then for all x, |x − c| < δ implies

|[2f(x) − 3g(x)] − [2f(c) − 3g(c)]| ≤ 2|f(x) − f(c)| + 3|g(x) − g(c)| < ϵ

Hence 2f(x) − 3g(x) is also continuous at x = c.

5. (a)∀ϵ > 0, ∃δ > 0 such that, for all x,

c − δ < x < c ⟹ |f(x) − L| < ϵ

(b)∀B > 0, ∃N < 0 such that, for all x,

x < N ⟹ f(x) > B