Calculus I, Fall 2014

Brief answers to Quiz 1

Oct 02, 2014

1. Give formal definition of lim_{x→c} f(x) ≠ L (just the definition, need not find ϵ or δ, etc.). Ans: There exists an ε_{0} > 0, such that for any δ > 0, ∃x_{0} ∈ (c − δ, c) ∪ (c, c + δ) and |f(x_{0}) − L| ≥ ε_{0}

2. Find lim_{θ→0} \frac{sin(sin θ)}{tan 2θ}.

Ans: 1/2.

3. State the Intermediate Value Theorem (Need not prove). Use it to find a c such that there is a root of ”x − 1 = cos x” on (c, c + 1). Ans: Let g(x) = x − 1 − cos x, then g is continuous on R. Since g(1) = − cos 1 < 0, g(2) = 1 − cos 2 > 0, by Intermediate Value Theorem, there exist x_{0} ∈ (1, 2) such that g(x_{0}) = 0, i.e. x_{0} − 1 = cos x_{0}. Therefore c = 1 works.

4. Use the ϵ − δ argument to show that if both f(x) and g(x) are continuous at x = c, then so is 2f(x) − 3g(x). Ans: 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}

Take 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. Give formal definitions of the following limits (Just the definition, need not find δ).

(a) lim_{x→c^{−}} f(x) = L (b) lim_{x→−∞} f(x) = ∞

Ans: (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