Calculus I, Fall 2021

Brief solutions to Quiz 6

Nov 30, 2021

1. (36 pts) Sketch the graph of f(x) = x^{4}−4x^{3}+10 (For your reference, f′(x) = 4x^{2}(x−3), f′′(x) = 12x(x − 2)) on [−1, 4]. Identify all local extremes and points of inflection, if any. Note that the domain is a finite closed interval. Ans: See page 3-5 of Lecture 15 note. Additional local maxima: f(−1) and f(4).

2. (16 pts) Evaluate lim_{x→0} \frac{1}{x} − \frac{1}{sin x}.

Ans: See Example 6 of Section 4.5.

3. (16 pts) Evaluate lim_{x→∞} (\frac{x + 1}{x − 1})^{x}.

Ans:

= lim_{x→∞} (1 + \frac{2}{x − 1})^{x} = e( lim_{x→∞}x ln(1 + \frac{2}{x − 1}))

lim_{x→∞}x ln(1 + \frac{2}{x − 1}) = lim_{x→∞} \frac{ln(1 + \frac{2}{x−1})}{\frac{1}{x}} = lim_{x→∞} ^{\frac{\frac{− \frac{2}{(x−1)^{2}}}{1+ \frac{2}{x−1}}}{\frac{−1}{x^{2}}}} = 2

Answer = e^{2}.

4. (16 pts) Evaluate lim_{x→0^{+}} x^{x}.

Ans: = e^{( lim_{x→0^{+}} x ln x)}

lim_{x→0^{+}} x ln x = lim_{x→0^{+}} \frac{ln x}{\frac{1}{x}} = lim_{x→0^{+}} \frac{\frac{1}{x}}{\frac{−1}{x^{2}}} = 0

Answer = e^{0} = 1.

5. (16 pts) Evaluate lim_{x→0^{+}} \frac{e^{\frac{−1}{x}}}{x}.

Ans: See page 6 of Lecture 16 note.